TPTP Problem File: NUM650^1.p
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% File : NUM650^1 : TPTP v9.0.0. Released v3.7.0.
% Domain : Number Theory
% Problem : Landau theorem 9a
% Version : Especial.
% English : ~(x = y) -> ~(~(forall x_0:nat.~(x = pl y x_0))) ->
% ~(forall x_0:nat.~(y = pl x x_0))
% Refs : [Lan30] Landau (1930), Grundlagen der Analysis
% : [vBJ79] van Benthem Jutting (1979), Checking Landau's "Grundla
% : [Bro09] Brown (2009), Email to Geoff Sutcliffe
% Source : [Bro09]
% Names : satz9a [Lan30]
% Status : Theorem
% : Without extensionality : Theorem
% Rating : 1.00 v3.7.0
% Syntax : Number of formulae : 20 ( 7 unt; 9 typ; 0 def)
% Number of atoms : 16 ( 10 equ; 0 cnn)
% Maximal formula atoms : 4 ( 1 avg)
% Number of connectives : 65 ( 11 ~; 0 |; 0 &; 45 @)
% ( 0 <=>; 9 =>; 0 <=; 0 <~>)
% Maximal formula depth : 9 ( 5 avg)
% Number of types : 3 ( 2 usr)
% Number of type conns : 9 ( 9 >; 0 *; 0 +; 0 <<)
% Number of symbols : 8 ( 7 usr; 3 con; 0-2 aty)
% Number of variables : 20 ( 0 ^; 20 !; 0 ?; 20 :)
% SPC : TH0_THM_EQU_NAR
% Comments :
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thf(nat_type,type,
nat: $tType ).
thf(x,type,
x: nat ).
thf(y,type,
y: nat ).
thf(pl,type,
pl: nat > nat > nat ).
thf(set_type,type,
set: $tType ).
thf(esti,type,
esti: nat > set > $o ).
thf(setof,type,
setof: ( nat > $o ) > set ).
thf(estie,axiom,
! [Xp: nat > $o,Xs: nat] :
( ( esti @ Xs @ ( setof @ Xp ) )
=> ( Xp @ Xs ) ) ).
thf(n_1,type,
n_1: nat ).
thf(suc,type,
suc: nat > nat ).
thf(ax5,axiom,
! [Xs: set] :
( ( esti @ n_1 @ Xs )
=> ( ! [Xx: nat] :
( ( esti @ Xx @ Xs )
=> ( esti @ ( suc @ Xx ) @ Xs ) )
=> ! [Xx: nat] : ( esti @ Xx @ Xs ) ) ) ).
thf(estii,axiom,
! [Xp: nat > $o,Xs: nat] :
( ( Xp @ Xs )
=> ( esti @ Xs @ ( setof @ Xp ) ) ) ).
thf(et,axiom,
! [Xa: $o] :
( ~ ~ Xa
=> Xa ) ).
thf(satz3,axiom,
! [Xx: nat] :
( ( Xx != n_1 )
=> ~ ! [Xx_0: nat] :
( Xx
!= ( suc @ Xx_0 ) ) ) ).
thf(satz4g,axiom,
! [Xx: nat] :
( ( suc @ Xx )
= ( pl @ n_1 @ Xx ) ) ).
thf(satz4e,axiom,
! [Xx: nat] :
( ( suc @ Xx )
= ( pl @ Xx @ n_1 ) ) ).
thf(satz4a,axiom,
! [Xx: nat] :
( ( pl @ Xx @ n_1 )
= ( suc @ Xx ) ) ).
thf(satz5,axiom,
! [Xx: nat,Xy: nat,Xz: nat] :
( ( pl @ ( pl @ Xx @ Xy ) @ Xz )
= ( pl @ Xx @ ( pl @ Xy @ Xz ) ) ) ).
thf(satz4f,axiom,
! [Xx: nat,Xy: nat] :
( ( suc @ ( pl @ Xx @ Xy ) )
= ( pl @ Xx @ ( suc @ Xy ) ) ) ).
thf(satz9a,conjecture,
( ( x != y )
=> ( ~ ~ ! [Xx_0: nat] :
( x
!= ( pl @ y @ Xx_0 ) )
=> ~ ! [Xx_0: nat] :
( y
!= ( pl @ x @ Xx_0 ) ) ) ) ).
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