TPTP Problem File: NUM696^1.p

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```%------------------------------------------------------------------------------
% File     : NUM696^1 : TPTP v7.2.0. Released v3.7.0.
% Domain   : Number Theory
% Problem  : Landau theorem 25
% Version  : Especial.
% English  : ~(~(forall x_0:nat.~(y = pl (pl x n_1) x_0))) -> y = pl x n_1

% 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    : satz25 [Lan30]

% Status   : Theorem
%          : Without extensionality : Theorem
% Rating   : 0.22 v7.2.0, 0.12 v7.1.0, 0.38 v7.0.0, 0.43 v6.4.0, 0.50 v6.3.0, 0.60 v6.2.0, 0.43 v5.5.0, 0.50 v5.4.0, 0.60 v5.1.0, 0.80 v5.0.0, 0.60 v4.1.0, 0.33 v4.0.1, 0.67 v4.0.0, 0.33 v3.7.0
% Syntax   : Number of formulae    :   11 (   0 unit;   5 type;   0 defn)
%            Number of atoms       :   48 (   8 equality;  19 variable)
%            Maximal formula depth :   13 (   6 average)
%            Number of connectives :   40 (  14   ~;   0   |;   0   &;  22   @)
%                                         (   0 <=>;   4  =>;   0  <=;   0 <~>)
%                                         (   0  ~|;   0  ~&)
%            Number of type conns  :    2 (   2   >;   0   *;   0   +;   0  <<)
%            Number of symbols     :    7 (   5   :;   0   =)
%            Number of variables   :   12 (   0 sgn;  12   !;   0   ?;   0   ^)
%                                         (  12   :;   0  !>;   0  ?*)
%                                         (   0  @-;   0  @+)
% SPC      : TH0_THM_EQU_NAR

%------------------------------------------------------------------------------
thf(nat_type,type,(
nat: \$tType )).

thf(x,type,(
x: nat )).

thf(y,type,(
y: nat )).

thf(pl,type,(
pl: nat > nat > nat )).

thf(m,axiom,(
~ ( ! [Xx_0: nat] :
( y
!= ( pl @ x @ Xx_0 ) ) ) )).

thf(n_1,type,(
n_1: nat )).

thf(et,axiom,(
! [Xa: \$o] :
( ~ ( ~ ( Xa ) )
=> Xa ) )).

thf(satz24,axiom,(
! [Xx: nat] :
( ~ ( ~ ( ! [Xx_0: nat] :
( Xx
!= ( pl @ n_1 @ Xx_0 ) ) ) )
=> ( Xx = n_1 ) ) )).

thf(satz19a,axiom,(
! [Xx: nat,Xy: nat,Xz: nat] :
( ~ ( ! [Xx_0: nat] :
( Xx
!= ( pl @ Xy @ Xx_0 ) ) )
=> ~ ( ! [Xx_0: nat] :
( ( pl @ Xx @ Xz )
!= ( pl @ ( pl @ Xy @ Xz ) @ Xx_0 ) ) ) ) )).

thf(satz6,axiom,(
! [Xx: nat,Xy: nat] :
( ( pl @ Xx @ Xy )
= ( pl @ Xy @ Xx ) ) )).

thf(satz25,conjecture,
( ~ ( ~ ( ! [Xx_0: nat] :
( y
!= ( pl @ ( pl @ x @ n_1 ) @ Xx_0 ) ) ) )
=> ( y
= ( pl @ x @ n_1 ) ) )).

%------------------------------------------------------------------------------
```