# Theory Classical

Up to index of Isabelle/HOL/ex

theory Classical
imports Main
begin

    (*  Title:      HOL/ex/Classical
        ID:         $Id: Classical.thy,v 1.10 2005/06/28 15:56:04 paulson Exp $
        Author:     Lawrence C Paulson, Cambridge University Computer Laboratory
        Copyright   1994  University of Cambridge
    *)

    header{*Classical Predicate Calculus Problems*}

    theory Classical imports Main begin

    subsection{*Traditional Classical Reasoner*}

    text{*The machine "griffon" mentioned below is a 2.5GHz Power Mac G5.*}

    text{*Taken from @{text "FOL/Classical.thy"}. When porting examples from
    first-order logic, beware of the precedence of @{text "="} versus @{text
    "\<leftrightarrow>"}.*}

    lemma "(P --> Q | R) --> (P-->Q) | (P-->R)"
    by blast

    text{*If and only if*}

    lemma "(P=Q) = (Q = (P::bool))"
    by blast

    lemma "~ (P = (~P))"
    by blast

    text{*Sample problems from
      F. J. Pelletier,
      Seventy-Five Problems for Testing Automatic Theorem Provers,
      J. Automated Reasoning 2 (1986), 191-216.
      Errata, JAR 4 (1988), 236-236.

    The hardest problems -- judging by experience with several theorem provers,
    including matrix ones -- are 34 and 43.
    *}

    subsubsection{*Pelletier's examples*}

    text{*1*}
    lemma "(P-->Q)  =  (~Q --> ~P)"
    by blast

    text{*2*}
    lemma "(~ ~ P) =  P"
    by blast

    text{*3*}
    lemma "~(P-->Q) --> (Q-->P)"
    by blast

    text{*4*}
    lemma "(~P-->Q)  =  (~Q --> P)"
    by blast

    text{*5*}
    lemma "((P|Q)-->(P|R)) --> (P|(Q-->R))"
    by blast

    text{*6*}
    lemma "P | ~ P"
    by blast

    text{*7*}
    lemma "P | ~ ~ ~ P"
    by blast

    text{*8.  Peirce's law*}
    lemma "((P-->Q) --> P)  -->  P"
    by blast

    text{*9*}
    lemma "((P|Q) & (~P|Q) & (P| ~Q)) --> ~ (~P | ~Q)"
    by blast

    text{*10*}
    lemma "(Q-->R) & (R-->P&Q) & (P-->Q|R) --> (P=Q)"
    by blast

    text{*11.  Proved in each direction (incorrectly, says Pelletier!!)  *}
    lemma "P=(P::bool)"
    by blast

    text{*12.  "Dijkstra's law"*}
    lemma "((P = Q) = R) = (P = (Q = R))"
    by blast

    text{*13.  Distributive law*}
    lemma "(P | (Q & R)) = ((P | Q) & (P | R))"
    by blast

    text{*14*}
    lemma "(P = Q) = ((Q | ~P) & (~Q|P))"
    by blast

    text{*15*}
    lemma "(P --> Q) = (~P | Q)"
    by blast

    text{*16*}
    lemma "(P-->Q) | (Q-->P)"
    by blast

    text{*17*}
    lemma "((P & (Q-->R))-->S)  =  ((~P | Q | S) & (~P | ~R | S))"
    by blast

    subsubsection{*Classical Logic: examples with quantifiers*}

    lemma "(∀x. P(x) & Q(x)) = ((∀x. P(x)) & (∀x. Q(x)))"
    by blast

    lemma "(∃x. P-->Q(x))  =  (P --> (∃x. Q(x)))"
    by blast

    lemma "(∃x. P(x)-->Q) = ((∀x. P(x)) --> Q)"
    by blast

    lemma "((∀x. P(x)) | Q)  =  (∀x. P(x) | Q)"
    by blast

    text{*From Wishnu Prasetya*}
    lemma "(∀s. q(s) --> r(s)) & ~r(s) & (∀s. ~r(s) & ~q(s) --> p(t) | q(t))
        --> p(t) | r(t)"
    by blast

    subsubsection{*Problems requiring quantifier duplication*}

    text{*Theorem B of Peter Andrews, Theorem Proving via General Matings,
      JACM 28 (1981).*}
    lemma "(∃x. ∀y. P(x) = P(y)) --> ((∃x. P(x)) = (∀y. P(y)))"
    by blast

    text{*Needs multiple instantiation of the quantifier.*}
    lemma "(∀x. P(x)-->P(f(x)))  &  P(d)-->P(f(f(f(d))))"
    by blast

    text{*Needs double instantiation of the quantifier*}
    lemma "∃x. P(x) --> P(a) & P(b)"
    by blast

    lemma "∃z. P(z) --> (∀x. P(x))"
    by blast

    lemma "∃x. (∃y. P(y)) --> P(x)"
    by blast

    subsubsection{*Hard examples with quantifiers*}

    text{*Problem 18*}
    lemma "∃y. ∀x. P(y)-->P(x)"
    by blast

    text{*Problem 19*}
    lemma "∃x. ∀y z. (P(y)-->Q(z)) --> (P(x)-->Q(x))"
    by blast

    text{*Problem 20*}
    lemma "(∀x y. ∃z. ∀w. (P(x)&Q(y)-->R(z)&S(w)))
        --> (∃x y. P(x) & Q(y)) --> (∃z. R(z))"
    by blast

    text{*Problem 21*}
    lemma "(∃x. P-->Q(x)) & (∃x. Q(x)-->P) --> (∃x. P=Q(x))"
    by blast

    text{*Problem 22*}
    lemma "(∀x. P = Q(x))  -->  (P = (∀x. Q(x)))"
    by blast

    text{*Problem 23*}
    lemma "(∀x. P | Q(x))  =  (P | (∀x. Q(x)))"
    by blast

    text{*Problem 24*}
    lemma "~(∃x. S(x)&Q(x)) & (∀x. P(x) --> Q(x)|R(x)) &
         (~(∃x. P(x)) --> (∃x. Q(x))) & (∀x. Q(x)|R(x) --> S(x))
        --> (∃x. P(x)&R(x))"
    by blast

    text{*Problem 25*}
    lemma "(∃x. P(x)) &
            (∀x. L(x) --> ~ (M(x) & R(x))) &
            (∀x. P(x) --> (M(x) & L(x))) &
            ((∀x. P(x)-->Q(x)) | (∃x. P(x)&R(x)))
        --> (∃x. Q(x)&P(x))"
    by blast

    text{*Problem 26*}
    lemma "((∃x. p(x)) = (∃x. q(x))) &
          (∀x. ∀y. p(x) & q(y) --> (r(x) = s(y)))
      --> ((∀x. p(x)-->r(x)) = (∀x. q(x)-->s(x)))"
    by blast

    text{*Problem 27*}
    lemma "(∃x. P(x) & ~Q(x)) &
                  (∀x. P(x) --> R(x)) &
                  (∀x. M(x) & L(x) --> P(x)) &
                  ((∃x. R(x) & ~ Q(x)) --> (∀x. L(x) --> ~ R(x)))
              --> (∀x. M(x) --> ~L(x))"
    by blast

    text{*Problem 28.  AMENDED*}
    lemma "(∀x. P(x) --> (∀x. Q(x))) &
            ((∀x. Q(x)|R(x)) --> (∃x. Q(x)&S(x))) &
            ((∃x. S(x)) --> (∀x. L(x) --> M(x)))
        --> (∀x. P(x) & L(x) --> M(x))"
    by blast

    text{*Problem 29.  Essentially the same as Principia Mathematica *11.71*}
    lemma "(∃x. F(x)) & (∃y. G(y))
        --> ( ((∀x. F(x)-->H(x)) & (∀y. G(y)-->J(y)))  =
              (∀x y. F(x) & G(y) --> H(x) & J(y)))"
    by blast

    text{*Problem 30*}
    lemma "(∀x. P(x) | Q(x) --> ~ R(x)) &
            (∀x. (Q(x) --> ~ S(x)) --> P(x) & R(x))
        --> (∀x. S(x))"
    by blast

    text{*Problem 31*}
    lemma "~(∃x. P(x) & (Q(x) | R(x))) &
            (∃x. L(x) & P(x)) &
            (∀x. ~ R(x) --> M(x))
        --> (∃x. L(x) & M(x))"
    by blast

    text{*Problem 32*}
    lemma "(∀x. P(x) & (Q(x)|R(x))-->S(x)) &
            (∀x. S(x) & R(x) --> L(x)) &
            (∀x. M(x) --> R(x))
        --> (∀x. P(x) & M(x) --> L(x))"
    by blast

    text{*Problem 33*}
    lemma "(∀x. P(a) & (P(x)-->P(b))-->P(c))  =
         (∀x. (~P(a) | P(x) | P(c)) & (~P(a) | ~P(b) | P(c)))"
    by blast

    text{*Problem 34  AMENDED (TWICE!!)*}
    text{*Andrews's challenge*}
    lemma "((∃x. ∀y. p(x) = p(y))  =
                   ((∃x. q(x)) = (∀y. p(y))))   =
                  ((∃x. ∀y. q(x) = q(y))  =
                   ((∃x. p(x)) = (∀y. q(y))))"
    by blast

    text{*Problem 35*}
    lemma "∃x y. P x y -->  (∀u v. P u v)"
    by blast

    text{*Problem 36*}
    lemma "(∀x. ∃y. J x y) &
            (∀x. ∃y. G x y) &
            (∀x y. J x y | G x y -->
            (∀z. J y z | G y z --> H x z))
        --> (∀x. ∃y. H x y)"
    by blast

    text{*Problem 37*}
    lemma "(∀z. ∃w. ∀x. ∃y.
               (P x z -->P y w) & P y z & (P y w --> (∃u. Q u w))) &
            (∀x z. ~(P x z) --> (∃y. Q y z)) &
            ((∃x y. Q x y) --> (∀x. R x x))
        --> (∀x. ∃y. R x y)"
    by blast

    text{*Problem 38*}
    lemma "(∀x. p(a) & (p(x) --> (∃y. p(y) & r x y)) -->
               (∃z. ∃w. p(z) & r x w & r w z))  =
         (∀x. (~p(a) | p(x) | (∃z. ∃w. p(z) & r x w & r w z)) &
               (~p(a) | ~(∃y. p(y) & r x y) |
                (∃z. ∃w. p(z) & r x w & r w z)))"
    by blast (*beats fast!*)

    text{*Problem 39*}
    lemma "~ (∃x. ∀y. F y x = (~ F y y))"
    by blast

    text{*Problem 40.  AMENDED*}
    lemma "(∃y. ∀x. F x y = F x x)
            -->  ~ (∀x. ∃y. ∀z. F z y = (~ F z x))"
    by blast

    text{*Problem 41*}
    lemma "(∀z. ∃y. ∀x. f x y = (f x z & ~ f x x))
                   --> ~ (∃z. ∀x. f x z)"
    by blast

    text{*Problem 42*}
    lemma "~ (∃y. ∀x. p x y = (~ (∃z. p x z & p z x)))"
    by blast

    text{*Problem 43!!*}
    lemma "(∀x::'a. ∀y::'a. q x y = (∀z. p z x = (p z y::bool)))
      --> (∀x. (∀y. q x y = (q y x::bool)))"
    by blast

    text{*Problem 44*}
    lemma "(∀x. f(x) -->
                  (∃y. g(y) & h x y & (∃y. g(y) & ~ h x y)))  &
                  (∃x. j(x) & (∀y. g(y) --> h x y))
                  --> (∃x. j(x) & ~f(x))"
    by blast

    text{*Problem 45*}
    lemma "(∀x. f(x) & (∀y. g(y) & h x y --> j x y)
                          --> (∀y. g(y) & h x y --> k(y))) &
         ~ (∃y. l(y) & k(y)) &
         (∃x. f(x) & (∀y. h x y --> l(y))
                    & (∀y. g(y) & h x y --> j x y))
          --> (∃x. f(x) & ~ (∃y. g(y) & h x y))"
    by blast

    subsubsection{*Problems (mainly) involving equality or functions*}

    text{*Problem 48*}
    lemma "(a=b | c=d) & (a=c | b=d) --> a=d | b=c"
    by blast

    text{*Problem 49  NOT PROVED AUTOMATICALLY.
         Hard because it involves substitution for Vars
      the type constraint ensures that x,y,z have the same type as a,b,u. *}
    lemma "(∃x y::'a. ∀z. z=x | z=y) & P(a) & P(b) & (~a=b)
                    --> (∀u::'a. P(u))"
    apply safe
    apply (rule_tac x = a in allE, assumption)
    apply (rule_tac x = b in allE, assumption, fast)  --{*blast's treatment of equality can't do it*}
    done

    text{*Problem 50.  (What has this to do with equality?) *}
    lemma "(∀x. P a x | (∀y. P x y)) --> (∃x. ∀y. P x y)"
    by blast

    text{*Problem 51*}
    lemma "(∃z w. ∀x y. P x y = (x=z & y=w)) -->
         (∃z. ∀x. ∃w. (∀y. P x y = (y=w)) = (x=z))"
    by blast

    text{*Problem 52. Almost the same as 51. *}
    lemma "(∃z w. ∀x y. P x y = (x=z & y=w)) -->
         (∃w. ∀y. ∃z. (∀x. P x y = (x=z)) = (y=w))"
    by blast

    text{*Problem 55*}

    text{*Non-equational version, from Manthey and Bry, CADE-9 (Springer, 1988).
      fast DISCOVERS who killed Agatha. *}
    lemma "lives(agatha) & lives(butler) & lives(charles) &
       (killed agatha agatha | killed butler agatha | killed charles agatha) &
       (∀x y. killed x y --> hates x y & ~richer x y) &
       (∀x. hates agatha x --> ~hates charles x) &
       (hates agatha agatha & hates agatha charles) &
       (∀x. lives(x) & ~richer x agatha --> hates butler x) &
       (∀x. hates agatha x --> hates butler x) &
       (∀x. ~hates x agatha | ~hates x butler | ~hates x charles) -->
        killed ?who agatha"
    by fast

    text{*Problem 56*}
    lemma "(∀x. (∃y. P(y) & x=f(y)) --> P(x)) = (∀x. P(x) --> P(f(x)))"
    by blast

    text{*Problem 57*}
    lemma "P (f a b) (f b c) & P (f b c) (f a c) &
         (∀x y z. P x y & P y z --> P x z)    -->   P (f a b) (f a c)"
    by blast

    text{*Problem 58  NOT PROVED AUTOMATICALLY*}
    lemma "(∀x y. f(x)=g(y)) --> (∀x y. f(f(x))=f(g(y)))"
    by (fast intro: arg_cong [of concl: f])

    text{*Problem 59*}
    lemma "(∀x. P(x) = (~P(f(x)))) --> (∃x. P(x) & ~P(f(x)))"
    by blast

    text{*Problem 60*}
    lemma "∀x. P x (f x) = (∃y. (∀z. P z y --> P z (f x)) & P x y)"
    by blast

    text{*Problem 62 as corrected in JAR 18 (1997), page 135*}
    lemma "(∀x. p a & (p x --> p(f x)) --> p(f(f x)))  =
          (∀x. (~ p a | p x | p(f(f x))) &
                  (~ p a | ~ p(f x) | p(f(f x))))"
    by blast

    text{*From Davis, Obvious Logical Inferences, IJCAI-81, 530-531
      fast indeed copes!*}
    lemma "(∀x. F(x) & ~G(x) --> (∃y. H(x,y) & J(y))) &
           (∃x. K(x) & F(x) & (∀y. H(x,y) --> K(y))) &
           (∀x. K(x) --> ~G(x))  -->  (∃x. K(x) & J(x))"
    by fast

    text{*From Rudnicki, Obvious Inferences, JAR 3 (1987), 383-393.
      It does seem obvious!*}
    lemma "(∀x. F(x) & ~G(x) --> (∃y. H(x,y) & J(y))) &
           (∃x. K(x) & F(x) & (∀y. H(x,y) --> K(y)))  &
           (∀x. K(x) --> ~G(x))   -->   (∃x. K(x) --> ~G(x))"
    by fast

    text{*Attributed to Lewis Carroll by S. G. Pulman.  The first or last
    assumption can be deleted.*}
    lemma "(∀x. honest(x) & industrious(x) --> healthy(x)) &
          ~ (∃x. grocer(x) & healthy(x)) &
          (∀x. industrious(x) & grocer(x) --> honest(x)) &
          (∀x. cyclist(x) --> industrious(x)) &
          (∀x. ~healthy(x) & cyclist(x) --> ~honest(x))
          --> (∀x. grocer(x) --> ~cyclist(x))"
    by blast

    lemma "(∀x y. R(x,y) | R(y,x)) &
           (∀x y. S(x,y) & S(y,x) --> x=y) &
           (∀x y. R(x,y) --> S(x,y))    -->   (∀x y. S(x,y) --> R(x,y))"
    by blast

    subsection{*Model Elimination Prover*}

    text{*Trying out meson with arguments*}
    lemma "x < y & y < z --> ~ (z < (x::nat))"
    by (meson order_less_irrefl order_less_trans)

    text{*The "small example" from Bezem, Hendriks and de Nivelle,
    Automatic Proof Construction in Type Theory Using Resolution,
    JAR 29: 3-4 (2002), pages 253-275 *}
    lemma "(∀x y z. R(x,y) & R(y,z) --> R(x,z)) &
           (∀x. ∃y. R(x,y)) -->
           ~ (∀x. P x = (∀y. R(x,y) --> ~ P y))"
    by (tactic{*safe_best_meson_tac 1*})
        --{*In contrast, @{text meson} is SLOW: 7.6s on griffon*}

    subsubsection{*Pelletier's examples*}
    text{*1*}
    lemma "(P --> Q)  =  (~Q --> ~P)"
    by blast

    text{*2*}
    lemma "(~ ~ P) =  P"
    by blast

    text{*3*}
    lemma "~(P-->Q) --> (Q-->P)"
    by blast

    text{*4*}
    lemma "(~P-->Q)  =  (~Q --> P)"
    by blast

    text{*5*}
    lemma "((P|Q)-->(P|R)) --> (P|(Q-->R))"
    by blast

    text{*6*}
    lemma "P | ~ P"
    by blast

    text{*7*}
    lemma "P | ~ ~ ~ P"
    by blast

    text{*8.  Peirce's law*}
    lemma "((P-->Q) --> P)  -->  P"
    by blast

    text{*9*}
    lemma "((P|Q) & (~P|Q) & (P| ~Q)) --> ~ (~P | ~Q)"
    by blast

    text{*10*}
    lemma "(Q-->R) & (R-->P&Q) & (P-->Q|R) --> (P=Q)"
    by blast

    text{*11.  Proved in each direction (incorrectly, says Pelletier!!)  *}
    lemma "P=(P::bool)"
    by blast

    text{*12.  "Dijkstra's law"*}
    lemma "((P = Q) = R) = (P = (Q = R))"
    by blast

    text{*13.  Distributive law*}
    lemma "(P | (Q & R)) = ((P | Q) & (P | R))"
    by blast

    text{*14*}
    lemma "(P = Q) = ((Q | ~P) & (~Q|P))"
    by blast

    text{*15*}
    lemma "(P --> Q) = (~P | Q)"
    by blast

    text{*16*}
    lemma "(P-->Q) | (Q-->P)"
    by blast

    text{*17*}
    lemma "((P & (Q-->R))-->S)  =  ((~P | Q | S) & (~P | ~R | S))"
    by blast

    subsubsection{*Classical Logic: examples with quantifiers*}

    lemma "(∀x. P x & Q x) = ((∀x. P x) & (∀x. Q x))"
    by blast

    lemma "(∃x. P --> Q x)  =  (P --> (∃x. Q x))"
    by blast

    lemma "(∃x. P x --> Q) = ((∀x. P x) --> Q)"
    by blast

    lemma "((∀x. P x) | Q)  =  (∀x. P x | Q)"
    by blast

    lemma "(∀x. P x --> P(f x))  &  P d --> P(f(f(f d)))"
    by blast

    text{*Needs double instantiation of EXISTS*}
    lemma "∃x. P x --> P a & P b"
    by blast

    lemma "∃z. P z --> (∀x. P x)"
    by blast

    text{*From a paper by Claire Quigley*}
    lemma "∃y. ((P c & Q y) | (∃z. ~ Q z)) | (∃x. ~ P x & Q d)"
    by fast

    subsubsection{*Hard examples with quantifiers*}

    text{*Problem 18*}
    lemma "∃y. ∀x. P y --> P x"
    by blast

    text{*Problem 19*}
    lemma "∃x. ∀y z. (P y --> Q z) --> (P x --> Q x)"
    by blast

    text{*Problem 20*}
    lemma "(∀x y. ∃z. ∀w. (P x & Q y --> R z & S w))
        --> (∃x y. P x & Q y) --> (∃z. R z)"
    by blast

    text{*Problem 21*}
    lemma "(∃x. P --> Q x) & (∃x. Q x --> P) --> (∃x. P=Q x)"
    by blast

    text{*Problem 22*}
    lemma "(∀x. P = Q x)  -->  (P = (∀x. Q x))"
    by blast

    text{*Problem 23*}
    lemma "(∀x. P | Q x)  =  (P | (∀x. Q x))"
    by blast

    text{*Problem 24*}  (*The first goal clause is useless*)
    lemma "~(∃x. S x & Q x) & (∀x. P x --> Q x | R x) &
          (~(∃x. P x) --> (∃x. Q x)) & (∀x. Q x | R x --> S x)
        --> (∃x. P x & R x)"
    by blast

    text{*Problem 25*}
    lemma "(∃x. P x) &
          (∀x. L x --> ~ (M x & R x)) &
          (∀x. P x --> (M x & L x)) &
          ((∀x. P x --> Q x) | (∃x. P x & R x))
        --> (∃x. Q x & P x)"
    by blast

    text{*Problem 26; has 24 Horn clauses*}
    lemma "((∃x. p x) = (∃x. q x)) &
          (∀x. ∀y. p x & q y --> (r x = s y))
      --> ((∀x. p x --> r x) = (∀x. q x --> s x))"
    by blast

    text{*Problem 27; has 13 Horn clauses*}
    lemma "(∃x. P x & ~Q x) &
          (∀x. P x --> R x) &
          (∀x. M x & L x --> P x) &
          ((∃x. R x & ~ Q x) --> (∀x. L x --> ~ R x))
          --> (∀x. M x --> ~L x)"
    by blast

    text{*Problem 28.  AMENDED; has 14 Horn clauses*}
    lemma "(∀x. P x --> (∀x. Q x)) &
          ((∀x. Q x | R x) --> (∃x. Q x & S x)) &
          ((∃x. S x) --> (∀x. L x --> M x))
        --> (∀x. P x & L x --> M x)"
    by blast

    text{*Problem 29.  Essentially the same as Principia Mathematica *11.71.
          62 Horn clauses*}
    lemma "(∃x. F x) & (∃y. G y)
        --> ( ((∀x. F x --> H x) & (∀y. G y --> J y))  =
              (∀x y. F x & G y --> H x & J y))"
    by blast

    text{*Problem 30*}
    lemma "(∀x. P x | Q x --> ~ R x) & (∀x. (Q x --> ~ S x) --> P x & R x)
           --> (∀x. S x)"
    by blast

    text{*Problem 31; has 10 Horn clauses; first negative clauses is useless*}
    lemma "~(∃x. P x & (Q x | R x)) &
          (∃x. L x & P x) &
          (∀x. ~ R x --> M x)
        --> (∃x. L x & M x)"
    by blast

    text{*Problem 32*}
    lemma "(∀x. P x & (Q x | R x)-->S x) &
          (∀x. S x & R x --> L x) &
          (∀x. M x --> R x)
        --> (∀x. P x & M x --> L x)"
    by blast

    text{*Problem 33; has 55 Horn clauses*}
    lemma "(∀x. P a & (P x --> P b)-->P c)  =
          (∀x. (~P a | P x | P c) & (~P a | ~P b | P c))"
    by blast

    text{*Problem 34: Andrews's challenge has 924 Horn clauses*}
    lemma "((∃x. ∀y. p x = p y)  = ((∃x. q x) = (∀y. p y)))     =
          ((∃x. ∀y. q x = q y)  = ((∃x. p x) = (∀y. q y)))"
    by blast

    text{*Problem 35*}
    lemma "∃x y. P x y -->  (∀u v. P u v)"
    by blast

    text{*Problem 36; has 15 Horn clauses*}
    lemma "(∀x. ∃y. J x y) & (∀x. ∃y. G x y) &
           (∀x y. J x y | G x y --> (∀z. J y z | G y z --> H x z))
           --> (∀x. ∃y. H x y)"
    by blast

    text{*Problem 37; has 10 Horn clauses*}
    lemma "(∀z. ∃w. ∀x. ∃y.
               (P x z --> P y w) & P y z & (P y w --> (∃u. Q u w))) &
          (∀x z. ~P x z --> (∃y. Q y z)) &
          ((∃x y. Q x y) --> (∀x. R x x))
        --> (∀x. ∃y. R x y)"
    by blast --{*causes unification tracing messages*}

    text{*Problem 38*}  text{*Quite hard: 422 Horn clauses!!*}
    lemma "(∀x. p a & (p x --> (∃y. p y & r x y)) -->
               (∃z. ∃w. p z & r x w & r w z))  =
          (∀x. (~p a | p x | (∃z. ∃w. p z & r x w & r w z)) &
                (~p a | ~(∃y. p y & r x y) |
                 (∃z. ∃w. p z & r x w & r w z)))"
    by blast

    text{*Problem 39*}
    lemma "~ (∃x. ∀y. F y x = (~F y y))"
    by blast

    text{*Problem 40.  AMENDED*}
    lemma "(∃y. ∀x. F x y = F x x)
          -->  ~ (∀x. ∃y. ∀z. F z y = (~F z x))"
    by blast

    text{*Problem 41*}
    lemma "(∀z. (∃y. (∀x. f x y = (f x z & ~ f x x))))
          --> ~ (∃z. ∀x. f x z)"
    by blast

    text{*Problem 42*}
    lemma "~ (∃y. ∀x. p x y = (~ (∃z. p x z & p z x)))"
    by blast

    text{*Problem 43  NOW PROVED AUTOMATICALLY!!*}
    lemma "(∀x. ∀y. q x y = (∀z. p z x = (p z y::bool)))
          --> (∀x. (∀y. q x y = (q y x::bool)))"
    by blast

    text{*Problem 44: 13 Horn clauses; 7-step proof*}
    lemma "(∀x. f x --> (∃y. g y & h x y & (∃y. g y & ~ h x y)))  &
           (∃x. j x & (∀y. g y --> h x y))
           --> (∃x. j x & ~f x)"
    by blast

    text{*Problem 45; has 27 Horn clauses; 54-step proof*}
    lemma "(∀x. f x & (∀y. g y & h x y --> j x y)
                --> (∀y. g y & h x y --> k y)) &
          ~ (∃y. l y & k y) &
          (∃x. f x & (∀y. h x y --> l y)
                    & (∀y. g y & h x y --> j x y))
          --> (∃x. f x & ~ (∃y. g y & h x y))"
    by blast

    text{*Problem 46; has 26 Horn clauses; 21-step proof*}
    lemma "(∀x. f x & (∀y. f y & h y x --> g y) --> g x) &
           ((∃x. f x & ~g x) -->
           (∃x. f x & ~g x & (∀y. f y & ~g y --> j x y))) &
           (∀x y. f x & f y & h x y --> ~j y x)
           --> (∀x. f x --> g x)"
    by blast

    text{*Problem 47.  Schubert's Steamroller.
          26 clauses; 63 Horn clauses.
          87094 inferences so far.  Searching to depth 36*}
    lemma "(∀x. wolf x --> animal x) & (∃x. wolf x) &
           (∀x. fox x --> animal x) & (∃x. fox x) &
           (∀x. bird x --> animal x) & (∃x. bird x) &
           (∀x. caterpillar x --> animal x) & (∃x. caterpillar x) &
           (∀x. snail x --> animal x) & (∃x. snail x) &
           (∀x. grain x --> plant x) & (∃x. grain x) &
           (∀x. animal x -->
                 ((∀y. plant y --> eats x y)  ∨
                  (∀y. animal y & smaller_than y x &
                        (∃z. plant z & eats y z) --> eats x y))) &
           (∀x y. bird y & (snail x ∨ caterpillar x) --> smaller_than x y) &
           (∀x y. bird x & fox y --> smaller_than x y) &
           (∀x y. fox x & wolf y --> smaller_than x y) &
           (∀x y. wolf x & (fox y ∨ grain y) --> ~eats x y) &
           (∀x y. bird x & caterpillar y --> eats x y) &
           (∀x y. bird x & snail y --> ~eats x y) &
           (∀x. (caterpillar x ∨ snail x) --> (∃y. plant y & eats x y))
           --> (∃x y. animal x & animal y & (∃z. grain z & eats y z & eats x y))"
    by (tactic{*safe_best_meson_tac 1*})
        --{*Nearly twice as fast as @{text meson},
            which performs iterative deepening rather than best-first search*}

    text{*The Los problem. Circulated by John Harrison*}
    lemma "(∀x y z. P x y & P y z --> P x z) &
           (∀x y z. Q x y & Q y z --> Q x z) &
           (∀x y. P x y --> P y x) &
           (∀x y. P x y | Q x y)
           --> (∀x y. P x y) | (∀x y. Q x y)"
    by meson

    text{*A similar example, suggested by Johannes Schumann and
     credited to Pelletier*}
    lemma "(∀x y z. P x y --> P y z --> P x z) -->
           (∀x y z. Q x y --> Q y z --> Q x z) -->
           (∀x y. Q x y --> Q y x) -->  (∀x y. P x y | Q x y) -->
           (∀x y. P x y) | (∀x y. Q x y)"
    by meson

    text{*Problem 50.  What has this to do with equality?*}
    lemma "(∀x. P a x | (∀y. P x y)) --> (∃x. ∀y. P x y)"
    by blast

    text{*Problem 54: NOT PROVED*}
    lemma "(∀y::'a. ∃z. ∀x. F x z = (x=y)) -->
          ~ (∃w. ∀x. F x w = (∀u. F x u --> (∃y. F y u & ~ (∃z. F z u & F z y))))"
    oops

    text{*Problem 55*}

    text{*Non-equational version, from Manthey and Bry, CADE-9 (Springer, 1988).
      @{text meson} cannot report who killed Agatha. *}
    lemma "lives agatha & lives butler & lives charles &
           (killed agatha agatha | killed butler agatha | killed charles agatha) &
           (∀x y. killed x y --> hates x y & ~richer x y) &
           (∀x. hates agatha x --> ~hates charles x) &
           (hates agatha agatha & hates agatha charles) &
           (∀x. lives x & ~richer x agatha --> hates butler x) &
           (∀x. hates agatha x --> hates butler x) &
           (∀x. ~hates x agatha | ~hates x butler | ~hates x charles) -->
           (∃x. killed x agatha)"
    by meson

    text{*Problem 57*}
    lemma "P (f a b) (f b c) & P (f b c) (f a c) &
          (∀x y z. P x y & P y z --> P x z)    -->   P (f a b) (f a c)"
    by blast

    text{*Problem 58: Challenge found on info-hol *}
    lemma "∀P Q R x. ∃v w. ∀y z. P x & Q y --> (P v | R w) & (R z --> Q v)"
    by blast

    text{*Problem 59*}
    lemma "(∀x. P x = (~P(f x))) --> (∃x. P x & ~P(f x))"
    by blast

    text{*Problem 60*}
    lemma "∀x. P x (f x) = (∃y. (∀z. P z y --> P z (f x)) & P x y)"
    by blast

    text{*Problem 62 as corrected in JAR 18 (1997), page 135*}
    lemma "(∀x. p a & (p x --> p(f x)) --> p(f(f x)))  =
           (∀x. (~ p a | p x | p(f(f x))) &
                (~ p a | ~ p(f x) | p(f(f x))))"
    by blast

    text{** Charles Morgan's problems **}

    lemma
      assumes a: "∀x y.  T(i x(i y x))"
          and b: "∀x y z. T(i (i x (i y z)) (i (i x y) (i x z)))"
          and c: "∀x y.   T(i (i (n x) (n y)) (i y x))"
          and c': "∀x y.   T(i (i y x) (i (n x) (n y)))"
          and d: "∀x y.   T(i x y) & T x --> T y"
     shows True
    proof -
      from a b d have "∀x. T(i x x)" by blast
      from a b c d have "∀x. T(i x (n(n x)))" --{*Problem 66*}
        by meson
          --{*SLOW: 18s on griffon. 208346 inferences, depth 23 *}
      from a b c d have "∀x. T(i (n(n x)) x)" --{*Problem 67*}
        by meson
          --{*4.9s on griffon. 51061 inferences, depth 21 *}
      from a b c' d have "∀x. T(i x (n(n x)))"
          --{*Problem 68: not proved.  Listed as satisfiable in TPTP (LCL078-1)*}
    oops

    text{*Problem 71, as found in TPTP (SYN007+1.005)*}
    lemma "p1 = (p2 = (p3 = (p4 = (p5 = (p1 = (p2 = (p3 = (p4 = p5))))))))"
    by blast

    text{*A manual resolution proof of problem 19.*}
    lemma "∃x. ∀y z. (P(y)-->Q(z)) --> (P(x)-->Q(x))"
    proof (rule ccontr, skolemize, make_clauses)
      fix f g
      assume P: "!!U. ¬ P U ==> False"
         and Q: "!!U. Q U ==> False"
         and PQ: "!!U.  [|P (f U); ¬ Q (g U)|] ==> False"
      have cl4: "!!U. ¬ Q (g U) ==> False"
        by (rule P [binary 0 PQ 0])
      show "False"
        by (rule Q [binary 0 cl4 0])
    qed

    end

### Traditional Classical Reasoner

lemma

      (P --> Q ∨ R) --> (P --> Q) ∨ (P --> R)

lemma

      (P = Q) = (Q = P)

lemma

      P ≠ (¬ P)

#### Pelletier's examples

lemma

      (P --> Q) = (¬ Q --> ¬ P)

lemma

      (¬ ¬ P) = P

lemma

      ¬ (P --> Q) --> Q --> P

lemma

      (¬ P --> Q) = (¬ Q --> P)

lemma

      (P ∨ Q --> P ∨ R) --> P ∨ (Q --> R)

lemma

      P ∨ ¬ P

lemma

      P ∨ ¬ ¬ ¬ P

lemma

      ((P --> Q) --> P) --> P

lemma

      (P ∨ Q) ∧ (¬ P ∨ Q) ∧ (P ∨ ¬ Q) --> ¬ (¬ P ∨ ¬ Q)

lemma

      (Q --> R) ∧ (R --> P ∧ Q) ∧ (P --> Q ∨ R) --> P = Q

lemma

      P = P

lemma

      ((P = Q) = R) = (P = (Q = R))

lemma

      (P ∨ Q ∧ R) = ((P ∨ Q) ∧ (P ∨ R))

lemma

      (P = Q) = ((Q ∨ ¬ P) ∧ (¬ Q ∨ P))

lemma

      (P --> Q) = (¬ P ∨ Q)

lemma

      (P --> Q) ∨ (Q --> P)

lemma

      (P ∧ (Q --> R) --> S) = ((¬ P ∨ Q ∨ S) ∧ (¬ P ∨ ¬ R ∨ S))

#### Classical Logic: examples with quantifiers

lemma

      (∀x. P x ∧ Q x) = ((∀x. P x) ∧ (∀x. Q x))

lemma

      (∃x. P --> Q x) = (P --> (∃x. Q x))

lemma

      (∃x. P x --> Q) = ((∀x. P x) --> Q)

lemma

      ((∀x. P x) ∨ Q) = (∀x. P x ∨ Q)

lemma

      (∀s. q s --> r s) ∧ ¬ r s ∧ (∀s. ¬ r s ∧ ¬ q s --> p t ∨ q t) --> p t ∨ r t

#### Problems requiring quantifier duplication

lemma

      (∃x. ∀y. P x = P y) --> (∃x. P x) = (∀y. P y)

lemma

      (∀x. P x --> P (f x)) ∧ P d --> P (f (f (f d)))

lemma

      ∃x. P x --> P a ∧ P b

lemma

      ∃z. P z --> (∀x. P x)

lemma

      ∃x. (∃y. P y) --> P x

#### Hard examples with quantifiers

lemma

      ∃y. ∀x. P y --> P x

lemma

      ∃x. ∀y z. (P y --> Q z) --> P x --> Q x

lemma

      (∀x y. ∃z. ∀w. P x ∧ Q y --> R z ∧ S w) --> (∃x y. P x ∧ Q y) --> (∃z. R z)

lemma

      (∃x. P --> Q x) ∧ (∃x. Q x --> P) --> (∃x. P = Q x)

lemma

      (∀x. P = Q x) --> P = (∀x. Q x)

lemma

      (∀x. P ∨ Q x) = (P ∨ (∀x. Q x))

lemma

      ¬ (∃x. S x ∧ Q x) ∧
      (∀x. P x --> Q x ∨ R x) ∧
      (¬ (∃x. P x) --> (∃x. Q x)) ∧ (∀x. Q x ∨ R x --> S x) -->
      (∃x. P x ∧ R x)

lemma

      (∃x. P x) ∧
      (∀x. L x --> ¬ (M x ∧ R x)) ∧
      (∀x. P x --> M x ∧ L x) ∧ ((∀x. P x --> Q x) ∨ (∃x. P x ∧ R x)) -->
      (∃x. Q x ∧ P x)

lemma

      (∃x. p x) = (∃x. q x) ∧ (∀x y. p x ∧ q y --> r x = s y) -->
      (∀x. p x --> r x) = (∀x. q x --> s x)

lemma

      (∃x. P x ∧ ¬ Q x) ∧
      (∀x. P x --> R x) ∧
      (∀x. M x ∧ L x --> P x) ∧ ((∃x. R x ∧ ¬ Q x) --> (∀x. L x --> ¬ R x)) -->
      (∀x. M x --> ¬ L x)

lemma

      (∀x. P x --> (∀x. Q x)) ∧
      ((∀x. Q x ∨ R x) --> (∃x. Q x ∧ S x)) ∧ ((∃x. S x) --> (∀x. L x --> M x)) -->
      (∀x. P x ∧ L x --> M x)

lemma

      (∃x. F x) ∧ (∃y. G y) -->
      ((∀x. F x --> H x) ∧ (∀y. G y --> J y)) = (∀x y. F x ∧ G y --> H x ∧ J y)

lemma

      (∀x. P x ∨ Q x --> ¬ R x) ∧ (∀x. (Q x --> ¬ S x) --> P x ∧ R x) --> (∀x. S x)

lemma

      ¬ (∃x. P x ∧ (Q x ∨ R x)) ∧ (∃x. L x ∧ P x) ∧ (∀x. ¬ R x --> M x) -->
      (∃x. L x ∧ M x)

lemma

      (∀x. P x ∧ (Q x ∨ R x) --> S x) ∧
      (∀x. S x ∧ R x --> L x) ∧ (∀x. M x --> R x) -->
      (∀x. P x ∧ M x --> L x)

lemma

      (∀x. P a ∧ (P x --> P b) --> P c) =
      (∀x. (¬ P a ∨ P x ∨ P c) ∧ (¬ P a ∨ ¬ P b ∨ P c))

lemma

      ((∃x. ∀y. p x = p y) = ((∃x. q x) = (∀y. p y))) =
      ((∃x. ∀y. q x = q y) = ((∃x. p x) = (∀y. q y)))

lemma

      ∃x y. P x y --> (∀u v. P u v)

lemma

      (∀x. ∃y. J x y) ∧
      (∀x. ∃y. G x y) ∧ (∀x y. J x y ∨ G x y --> (∀z. J y z ∨ G y z --> H x z)) -->
      (∀x. ∃y. H x y)

lemma

      (∀z. ∃w. ∀x. ∃y. (P x z --> P y w) ∧ P y z ∧ (P y w --> (∃u. Q u w))) ∧
      (∀x z. ¬ P x z --> (∃y. Q y z)) ∧ ((∃x y. Q x y) --> (∀x. R x x)) -->
      (∀x. ∃y. R x y)

lemma

      (∀x. p a ∧ (p x --> (∃y. p y ∧ r x y)) --> (∃z w. p z ∧ r x w ∧ r w z)) =
      (∀x. (¬ p a ∨ p x ∨ (∃z w. p z ∧ r x w ∧ r w z)) ∧
           (¬ p a ∨ ¬ (∃y. p y ∧ r x y) ∨ (∃z w. p z ∧ r x w ∧ r w z)))

lemma

      ¬ (∃x. ∀y. F y x = (¬ F y y))

lemma

      (∃y. ∀x. F x y = F x x) --> ¬ (∀x. ∃y. ∀z. F z y = (¬ F z x))

lemma

      (∀z. ∃y. ∀x. f x y = (f x z ∧ ¬ f x x)) --> ¬ (∃z. ∀x. f x z)

lemma

      ¬ (∃y. ∀x. p x y = (¬ (∃z. p x z ∧ p z x)))

lemma

      (∀x y. q x y = (∀z. p z x = p z y)) --> (∀x y. q x y = q y x)

lemma

      (∀x. f x --> (∃y. g y ∧ h x y ∧ (∃y. g y ∧ ¬ h x y))) ∧
      (∃x. j x ∧ (∀y. g y --> h x y)) -->
      (∃x. j x ∧ ¬ f x)

lemma

      (∀x. f x ∧ (∀y. g y ∧ h x y --> j x y) --> (∀y. g y ∧ h x y --> k y)) ∧
      ¬ (∃y. l y ∧ k y) ∧
      (∃x. f x ∧ (∀y. h x y --> l y) ∧ (∀y. g y ∧ h x y --> j x y)) -->
      (∃x. f x ∧ ¬ (∃y. g y ∧ h x y))

#### Problems (mainly) involving equality or functions

lemma

      (a = b ∨ c = d) ∧ (a = c ∨ b = d) --> a = d ∨ b = c

lemma

      (∃x y. ∀z. z = x ∨ z = y) ∧ P a ∧ P b ∧ a ≠ b --> (∀u. P u)

lemma

      (∀x. P a x ∨ (∀y. P x y)) --> (∃x. ∀y. P x y)

lemma

      (∃z w. ∀x y. P x y = (x = z ∧ y = w)) -->
      (∃z. ∀x. ∃w. (∀y. P x y = (y = w)) = (x = z))

lemma

      (∃z w. ∀x y. P x y = (x = z ∧ y = w)) -->
      (∃w. ∀y. ∃z. (∀x. P x y = (x = z)) = (y = w))

lemma

      lives agatha ∧
      lives butler ∧
      lives charles ∧
      (killed agatha agatha ∨ killed butler agatha ∨ killed charles agatha) ∧
      (∀x y. killed x y --> hates x y ∧ ¬ richer x y) ∧
      (∀x. hates agatha x --> ¬ hates charles x) ∧
      (hates agatha agatha ∧ hates agatha charles) ∧
      (∀x. lives x ∧ ¬ richer x agatha --> hates butler x) ∧
      (∀x. hates agatha x --> hates butler x) ∧
      (∀x. ¬ hates x agatha ∨ ¬ hates x butler ∨ ¬ hates x charles) -->
      killed agatha agatha

lemma

      (∀x. (∃y. P y ∧ x = f y) --> P x) = (∀x. P x --> P (f x))

lemma

      P (f a b) (f b c) ∧ P (f b c) (f a c) ∧ (∀x y z. P x y ∧ P y z --> P x z) -->
      P (f a b) (f a c)

lemma

      (∀x y. f x = g y) --> (∀x y. f (f x) = f (g y))

lemma

      (∀x. P x = (¬ P (f x))) --> (∃x. P x ∧ ¬ P (f x))

lemma

      ∀x. P x (f x) = (∃y. (∀z. P z y --> P z (f x)) ∧ P x y)

lemma

      (∀x. p a ∧ (p x --> p (f x)) --> p (f (f x))) =
      (∀x. (¬ p a ∨ p x ∨ p (f (f x))) ∧ (¬ p a ∨ ¬ p (f x) ∨ p (f (f x))))

lemma

      (∀x. F x ∧ ¬ G x --> (∃y. H (x, y) ∧ J y)) ∧
      (∃x. K x ∧ F x ∧ (∀y. H (x, y) --> K y)) ∧ (∀x. K x --> ¬ G x) -->
      (∃x. K x ∧ J x)

lemma

      (∀x. F x ∧ ¬ G x --> (∃y. H (x, y) ∧ J y)) ∧
      (∃x. K x ∧ F x ∧ (∀y. H (x, y) --> K y)) ∧ (∀x. K x --> ¬ G x) -->
      (∃x. K x --> ¬ G x)

lemma

      (∀x. honest x ∧ industrious x --> healthy x) ∧
      ¬ (∃x. grocer x ∧ healthy x) ∧
      (∀x. industrious x ∧ grocer x --> honest x) ∧
      (∀x. cyclist x --> industrious x) ∧
      (∀x. ¬ healthy x ∧ cyclist x --> ¬ honest x) -->
      (∀x. grocer x --> ¬ cyclist x)

lemma

      (∀x y. R (x, y) ∨ R (y, x)) ∧
      (∀x y. S (x, y) ∧ S (y, x) --> x = y) ∧ (∀x y. R (x, y) --> S (x, y)) -->
      (∀x y. S (x, y) --> R (x, y))

### Model Elimination Prover

lemma

      x < y ∧ y < z --> ¬ z < x

lemma

      (∀x y z. R (x, y) ∧ R (y, z) --> R (x, z)) ∧ (∀x. ∃y. R (x, y)) -->
      ¬ (∀x. P x = (∀y. R (x, y) --> ¬ P y))

#### Pelletier's examples

lemma

      (P --> Q) = (¬ Q --> ¬ P)

lemma

      (¬ ¬ P) = P

lemma

      ¬ (P --> Q) --> Q --> P

lemma

      (¬ P --> Q) = (¬ Q --> P)

lemma

      (P ∨ Q --> P ∨ R) --> P ∨ (Q --> R)

lemma

      P ∨ ¬ P

lemma

      P ∨ ¬ ¬ ¬ P

lemma

      ((P --> Q) --> P) --> P

lemma

      (P ∨ Q) ∧ (¬ P ∨ Q) ∧ (P ∨ ¬ Q) --> ¬ (¬ P ∨ ¬ Q)

lemma

      (Q --> R) ∧ (R --> P ∧ Q) ∧ (P --> Q ∨ R) --> P = Q

lemma

      P = P

lemma

      ((P = Q) = R) = (P = (Q = R))

lemma

      (P ∨ Q ∧ R) = ((P ∨ Q) ∧ (P ∨ R))

lemma

      (P = Q) = ((Q ∨ ¬ P) ∧ (¬ Q ∨ P))

lemma

      (P --> Q) = (¬ P ∨ Q)

lemma

      (P --> Q) ∨ (Q --> P)

lemma

      (P ∧ (Q --> R) --> S) = ((¬ P ∨ Q ∨ S) ∧ (¬ P ∨ ¬ R ∨ S))

#### Classical Logic: examples with quantifiers

lemma

      (∀x. P x ∧ Q x) = ((∀x. P x) ∧ (∀x. Q x))

lemma

      (∃x. P --> Q x) = (P --> (∃x. Q x))

lemma

      (∃x. P x --> Q) = ((∀x. P x) --> Q)

lemma

      ((∀x. P x) ∨ Q) = (∀x. P x ∨ Q)

lemma

      (∀x. P x --> P (f x)) ∧ P d --> P (f (f (f d)))

lemma

      ∃x. P x --> P a ∧ P b

lemma

      ∃z. P z --> (∀x. P x)

lemma

      ∃y. (P c ∧ Q y ∨ (∃z. ¬ Q z)) ∨ (∃x. ¬ P x ∧ Q d)

#### Hard examples with quantifiers

lemma

      ∃y. ∀x. P y --> P x

lemma

      ∃x. ∀y z. (P y --> Q z) --> P x --> Q x

lemma

      (∀x y. ∃z. ∀w. P x ∧ Q y --> R z ∧ S w) --> (∃x y. P x ∧ Q y) --> (∃z. R z)

lemma

      (∃x. P --> Q x) ∧ (∃x. Q x --> P) --> (∃x. P = Q x)

lemma

      (∀x. P = Q x) --> P = (∀x. Q x)

lemma

      (∀x. P ∨ Q x) = (P ∨ (∀x. Q x))

lemma

      ¬ (∃x. S x ∧ Q x) ∧
      (∀x. P x --> Q x ∨ R x) ∧
      (¬ (∃x. P x) --> (∃x. Q x)) ∧ (∀x. Q x ∨ R x --> S x) -->
      (∃x. P x ∧ R x)

lemma

      (∃x. P x) ∧
      (∀x. L x --> ¬ (M x ∧ R x)) ∧
      (∀x. P x --> M x ∧ L x) ∧ ((∀x. P x --> Q x) ∨ (∃x. P x ∧ R x)) -->
      (∃x. Q x ∧ P x)

lemma

      (∃x. p x) = (∃x. q x) ∧ (∀x y. p x ∧ q y --> r x = s y) -->
      (∀x. p x --> r x) = (∀x. q x --> s x)

lemma

      (∃x. P x ∧ ¬ Q x) ∧
      (∀x. P x --> R x) ∧
      (∀x. M x ∧ L x --> P x) ∧ ((∃x. R x ∧ ¬ Q x) --> (∀x. L x --> ¬ R x)) -->
      (∀x. M x --> ¬ L x)

lemma

      (∀x. P x --> (∀x. Q x)) ∧
      ((∀x. Q x ∨ R x) --> (∃x. Q x ∧ S x)) ∧ ((∃x. S x) --> (∀x. L x --> M x)) -->
      (∀x. P x ∧ L x --> M x)

lemma

      (∃x. F x) ∧ (∃y. G y) -->
      ((∀x. F x --> H x) ∧ (∀y. G y --> J y)) = (∀x y. F x ∧ G y --> H x ∧ J y)

lemma

      (∀x. P x ∨ Q x --> ¬ R x) ∧ (∀x. (Q x --> ¬ S x) --> P x ∧ R x) --> (∀x. S x)

lemma

      ¬ (∃x. P x ∧ (Q x ∨ R x)) ∧ (∃x. L x ∧ P x) ∧ (∀x. ¬ R x --> M x) -->
      (∃x. L x ∧ M x)

lemma

      (∀x. P x ∧ (Q x ∨ R x) --> S x) ∧
      (∀x. S x ∧ R x --> L x) ∧ (∀x. M x --> R x) -->
      (∀x. P x ∧ M x --> L x)

lemma

      (∀x. P a ∧ (P x --> P b) --> P c) =
      (∀x. (¬ P a ∨ P x ∨ P c) ∧ (¬ P a ∨ ¬ P b ∨ P c))

lemma

      ((∃x. ∀y. p x = p y) = ((∃x. q x) = (∀y. p y))) =
      ((∃x. ∀y. q x = q y) = ((∃x. p x) = (∀y. q y)))

lemma

      ∃x y. P x y --> (∀u v. P u v)

lemma

      (∀x. ∃y. J x y) ∧
      (∀x. ∃y. G x y) ∧ (∀x y. J x y ∨ G x y --> (∀z. J y z ∨ G y z --> H x z)) -->
      (∀x. ∃y. H x y)

lemma

      (∀z. ∃w. ∀x. ∃y. (P x z --> P y w) ∧ P y z ∧ (P y w --> (∃u. Q u w))) ∧
      (∀x z. ¬ P x z --> (∃y. Q y z)) ∧ ((∃x y. Q x y) --> (∀x. R x x)) -->
      (∀x. ∃y. R x y)

lemma

      (∀x. p a ∧ (p x --> (∃y. p y ∧ r x y)) --> (∃z w. p z ∧ r x w ∧ r w z)) =
      (∀x. (¬ p a ∨ p x ∨ (∃z w. p z ∧ r x w ∧ r w z)) ∧
           (¬ p a ∨ ¬ (∃y. p y ∧ r x y) ∨ (∃z w. p z ∧ r x w ∧ r w z)))

lemma

      ¬ (∃x. ∀y. F y x = (¬ F y y))

lemma

      (∃y. ∀x. F x y = F x x) --> ¬ (∀x. ∃y. ∀z. F z y = (¬ F z x))

lemma

      (∀z. ∃y. ∀x. f x y = (f x z ∧ ¬ f x x)) --> ¬ (∃z. ∀x. f x z)

lemma

      ¬ (∃y. ∀x. p x y = (¬ (∃z. p x z ∧ p z x)))

lemma

      (∀x y. q x y = (∀z. p z x = p z y)) --> (∀x y. q x y = q y x)

lemma

      (∀x. f x --> (∃y. g y ∧ h x y ∧ (∃y. g y ∧ ¬ h x y))) ∧
      (∃x. j x ∧ (∀y. g y --> h x y)) -->
      (∃x. j x ∧ ¬ f x)

lemma

      (∀x. f x ∧ (∀y. g y ∧ h x y --> j x y) --> (∀y. g y ∧ h x y --> k y)) ∧
      ¬ (∃y. l y ∧ k y) ∧
      (∃x. f x ∧ (∀y. h x y --> l y) ∧ (∀y. g y ∧ h x y --> j x y)) -->
      (∃x. f x ∧ ¬ (∃y. g y ∧ h x y))

lemma

      (∀x. f x ∧ (∀y. f y ∧ h y x --> g y) --> g x) ∧
      ((∃x. f x ∧ ¬ g x) --> (∃x. f x ∧ ¬ g x ∧ (∀y. f y ∧ ¬ g y --> j x y))) ∧
      (∀x y. f x ∧ f y ∧ h x y --> ¬ j y x) -->
      (∀x. f x --> g x)

lemma

      (∀x. wolf x --> animal x) ∧
      (∃x. wolf x) ∧
      (∀x. fox x --> animal x) ∧
      (∃x. fox x) ∧
      (∀x. bird x --> animal x) ∧
      (∃x. bird x) ∧
      (∀x. caterpillar x --> animal x) ∧
      (∃x. caterpillar x) ∧
      (∀x. snail x --> animal x) ∧
      (∃x. snail x) ∧
      (∀x. grain x --> plant x) ∧
      (∃x. grain x) ∧
      (∀x. animal x -->
           (∀y. plant y --> eats x y) ∨
           (∀y. animal y ∧ smaller_than y x ∧ (∃z. plant z ∧ eats y z) -->
                eats x y)) ∧
      (∀x y. bird y ∧ (snail x ∨ caterpillar x) --> smaller_than x y) ∧
      (∀x y. bird x ∧ fox y --> smaller_than x y) ∧
      (∀x y. fox x ∧ wolf y --> smaller_than x y) ∧
      (∀x y. wolf x ∧ (fox y ∨ grain y) --> ¬ eats x y) ∧
      (∀x y. bird x ∧ caterpillar y --> eats x y) ∧
      (∀x y. bird x ∧ snail y --> ¬ eats x y) ∧
      (∀x. caterpillar x ∨ snail x --> (∃y. plant y ∧ eats x y)) -->
      (∃x y. animal x ∧ animal y ∧ (∃z. grain z ∧ eats y z ∧ eats x y))

lemma

      (∀x y z. P x y ∧ P y z --> P x z) ∧
      (∀x y z. Q x y ∧ Q y z --> Q x z) ∧
      (∀x y. P x y --> P y x) ∧ (∀x y. P x y ∨ Q x y) -->
      (∀x y. P x y) ∨ (∀x y. Q x y)

lemma

      (∀x y z. P x y --> P y z --> P x z) -->
      (∀x y z. Q x y --> Q y z --> Q x z) -->
      (∀x y. Q x y --> Q y x) -->
      (∀x y. P x y ∨ Q x y) --> (∀x y. P x y) ∨ (∀x y. Q x y)

lemma

      (∀x. P a x ∨ (∀y. P x y)) --> (∃x. ∀y. P x y)

lemma

      lives agatha ∧
      lives butler ∧
      lives charles ∧
      (killed agatha agatha ∨ killed butler agatha ∨ killed charles agatha) ∧
      (∀x y. killed x y --> hates x y ∧ ¬ richer x y) ∧
      (∀x. hates agatha x --> ¬ hates charles x) ∧
      (hates agatha agatha ∧ hates agatha charles) ∧
      (∀x. lives x ∧ ¬ richer x agatha --> hates butler x) ∧
      (∀x. hates agatha x --> hates butler x) ∧
      (∀x. ¬ hates x agatha ∨ ¬ hates x butler ∨ ¬ hates x charles) -->
      (∃x. killed x agatha)

lemma

      P (f a b) (f b c) ∧ P (f b c) (f a c) ∧ (∀x y z. P x y ∧ P y z --> P x z) -->
      P (f a b) (f a c)

lemma

      ∀P Q R x. ∃v w. ∀y z. P x ∧ Q y --> (P v ∨ R w) ∧ (R z --> Q v)

lemma

      (∀x. P x = (¬ P (f x))) --> (∃x. P x ∧ ¬ P (f x))

lemma

      ∀x. P x (f x) = (∃y. (∀z. P z y --> P z (f x)) ∧ P x y)

lemma

      (∀x. p a ∧ (p x --> p (f x)) --> p (f (f x))) =
      (∀x. (¬ p a ∨ p x ∨ p (f (f x))) ∧ (¬ p a ∨ ¬ p (f x) ∨ p (f (f x))))

lemma

      p1.0 =
      (p2.0 = (p3.0 = (p4.0 = (p5.0 = (p1.0 = (p2.0 = (p3.0 = (p4.0 = p5.0))))))))

lemma

      ∃x. ∀y z. (P y --> Q z) --> P x --> Q x