TPTP Problem File: DAT358^1.p
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% File : DAT358^1 : TPTP v9.2.1. Released v9.2.0.
% Domain : Data Structures
% Problem : Associativity of list append, induction step 4
% Version : Especial.
% English :
% Refs : [RRB23] Rothgang et al. (2023), Theorem Proving in Dependently
% : [Rot25] Rothgang (2025), Email to Geoff Sutcliffe
% : [RK+25] Ranalter et al. (2025), The Dependently Typed Higher-O
% Source : [Rot25]
% Names : ListAppAssoc/list-app-assoc-indstep4.p [Rot25]
% Status : Theorem
% Rating : ? v9.2.0
% Syntax : Number of formulae : 13 ( 4 unt; 9 typ; 0 def)
% Number of atoms : 4 ( 4 equ; 0 cnn)
% Maximal formula atoms : 1 ( 1 avg)
% Number of connectives : 76 ( 0 ~; 0 |; 0 &; 76 @)
% ( 0 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 8 ( 5 avg)
% Number of types : 2 ( 2 usr)
% Number of type decls : 9 ( 0 !>P; 2 !>D)
% Number of type conns : 8 ( 8 >; 0 *; 0 +; 0 <<)
% Number of symbols : 8 ( 7 usr; 2 con; 0-4 aty)
% Number of variables : 20 ( 0 ^; 17 !; 0 ?; 20 :)
% ( 3 !>; 0 ?*; 0 @-; 0 @+)
% SPC : DH0_THM_EQU_NAR
% Comments :
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thf(elem_type,type,
elem: $tType ).
thf(nat_type,type,
nat: $tType ).
thf(zero_type,type,
zero: nat ).
thf(suc_type,type,
suc: nat > nat ).
thf(plus_type,type,
plus: nat > nat > nat ).
thf(list_type,type,
list: nat > $tType ).
thf(nil_type,type,
nil: list @ zero ).
thf(cons_type,type,
cons:
!>[N: nat] : ( elem > ( list @ N ) > ( list @ ( suc @ N ) ) ) ).
thf(app_type,type,
app:
!>[N: nat,M: nat] : ( ( list @ N ) > ( list @ M ) > ( list @ ( plus @ N @ M ) ) ) ).
thf(ax2,axiom,
! [N: nat,M: nat] :
( ( plus @ ( suc @ N ) @ M )
= ( suc @ ( plus @ N @ M ) ) ) ).
thf(ax4,axiom,
! [N: nat,M: nat,X: elem,Y: list @ N,Z: list @ M] :
( ( app @ ( suc @ N ) @ M @ ( cons @ N @ X @ Y ) @ Z )
= ( cons @ ( plus @ N @ M ) @ X @ ( app @ N @ M @ Y @ Z ) ) ) ).
thf(plus_assoc,axiom,
! [M1: nat,M2: nat,M3: nat] :
( ( plus @ M1 @ ( plus @ M2 @ M3 ) )
= ( plus @ ( plus @ M1 @ M2 ) @ M3 ) ) ).
thf(list_app_assoc_indstep4,conjecture,
! [M2: nat,L2: list @ M2,M3: nat,L3: list @ M3,M: nat,X: elem,L: list @ M] :
( ( app @ ( suc @ M ) @ ( plus @ M2 @ M3 ) @ ( cons @ M @ X @ L ) @ ( app @ M2 @ M3 @ L2 @ L3 ) )
= ( cons @ ( plus @ M @ ( plus @ M2 @ M3 ) ) @ X @ ( app @ M @ ( plus @ M2 @ M3 ) @ L @ ( app @ M2 @ M3 @ L2 @ L3 ) ) ) ) ).
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