TPTP Problem File: SYO038^1.004.004.p
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% File : SYO038^1.004.004 : TPTP v9.0.0. Released v5.3.0.
% Domain : Syntactic
% Problem : Boolos' Curious Inference, size f(4,4)
% Version : Especial.
% English :
% Refs : [Boo87] Boolos (1987), A Curious Inference
% : [BB05] Benzmueller & Brown (2005), A Structured Set of Higher
% : [BB07] Benzmueller & Brown (2007), The Curious Inference of B
% : [Ben09] Benzmueller (2009), Email to Geoff Sutcliffe
% Source : [Ben09]
% Names : Example 35 [BB05]
% Status : Theorem
% Rating : 1.00 v5.3.0
% Syntax : Number of formulae : 10 ( 5 unt; 4 typ; 0 def)
% Number of atoms : 7 ( 3 equ; 0 cnn)
% Maximal formula atoms : 2 ( 1 avg)
% Number of connectives : 33 ( 0 ~; 0 |; 0 &; 32 @)
% ( 0 <=>; 1 =>; 0 <=; 0 <~>)
% Maximal formula depth : 7 ( 4 avg)
% Number of types : 2 ( 0 usr)
% Number of type conns : 4 ( 4 >; 0 *; 0 +; 0 <<)
% Number of symbols : 5 ( 4 usr; 1 con; 0-2 aty)
% Number of variables : 5 ( 0 ^; 5 !; 0 ?; 5 :)
% SPC : TH0_THM_EQU_NAR
% Comments : The first order proof is infeasibly long - f(5,5) is huge. It is
% a number that makes googolplex look tiny. Practically impossible
% to solve using only first-order means. However, there exists a
% proof in higher-order logic that is very short but hard to find.
% : f(4,4) is 2^(2^(2...)) stacked 65536 times. It is huge. It is
% much bigger than googolplex.
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thf(one,type,
one: $i ).
thf(s,type,
s: $i > $i ).
thf(f,type,
f: $i > $i > $i ).
thf(d,type,
d: $i > $o ).
thf(ax1,axiom,
! [N: $i] :
( ( f @ N @ one )
= ( s @ one ) ) ).
thf(ax2,axiom,
! [X: $i] :
( ( f @ one @ ( s @ X ) )
= ( s @ ( s @ ( f @ one @ X ) ) ) ) ).
thf(ax3,axiom,
! [N: $i,X: $i] :
( ( f @ ( s @ N ) @ ( s @ X ) )
= ( f @ N @ ( f @ ( s @ N ) @ X ) ) ) ).
thf(ax4,axiom,
d @ one ).
thf(ax5,axiom,
! [X: $i] :
( ( d @ X )
=> ( d @ ( s @ X ) ) ) ).
thf(conj,conjecture,
d @ ( f @ ( s @ ( s @ ( s @ one ) ) ) @ ( s @ ( s @ ( s @ one ) ) ) ) ).
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