TSTP Solution File: ITP001+2 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : ITP001+2 : TPTP v8.1.0. Bugfixed v7.5.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n025.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Wed Jul 27 12:58:51 EDT 2022
% Result : Theorem 1.94s 2.13s
% Output : Refutation 1.94s
% Verified :
% SZS Type : Refutation
% Derivation depth : 1
% Number of leaves : 1
% Syntax : Number of clauses : 2 ( 2 unt; 0 nHn; 2 RR)
% Number of literals : 2 ( 0 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 0 ( 0 avg)
% Number of predicates : 2 ( 1 usr; 2 prp; 0-0 aty)
% Number of functors : 0 ( 0 usr; 0 con; --- aty)
% Number of variables : 0 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(14,axiom,
~ dollar_T,
file('ITP001+2.p',unknown),
[] ).
cnf(15,plain,
$false,
inference(propositional,[status(thm)],[inference(copy,[status(thm)],[14])]),
[iquote('copy,14,propositional')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : ITP001+2 : TPTP v8.1.0. Bugfixed v7.5.0.
% 0.11/0.13 % Command : otter-tptp-script %s
% 0.13/0.34 % Computer : n025.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Wed Jul 27 02:52:15 EDT 2022
% 0.13/0.34 % CPUTime :
% 1.94/2.13
% 1.94/2.13 -------- PROOF --------
% 1.94/2.13 ----- Otter 3.3f, August 2004 -----
% 1.94/2.13 The process was started by sandbox2 on n025.cluster.edu,
% 1.94/2.13 Wed Jul 27 02:52:16 2022
% 1.94/2.13 The command was "./otter". The process ID is 24116.
% 1.94/2.13
% 1.94/2.13 set(prolog_style_variables).
% 1.94/2.13 set(auto).
% 1.94/2.13 dependent: set(auto1).
% 1.94/2.13 dependent: set(process_input).
% 1.94/2.13 dependent: clear(print_kept).
% 1.94/2.13 dependent: clear(print_new_demod).
% 1.94/2.13 dependent: clear(print_back_demod).
% 1.94/2.13 dependent: clear(print_back_sub).
% 1.94/2.13 dependent: set(control_memory).
% 1.94/2.13 dependent: assign(max_mem, 12000).
% 1.94/2.13 dependent: assign(pick_given_ratio, 4).
% 1.94/2.13 dependent: assign(stats_level, 1).
% 1.94/2.13 dependent: assign(max_seconds, 10800).
% 1.94/2.13 clear(print_given).
% 1.94/2.13
% 1.94/2.13 formula_list(usable).
% 1.94/2.13 all A (A=A).
% 1.94/2.13 ne(bool).
% 1.94/2.13 ne(ind).
% 1.94/2.13 all A (ne(A)-> (all B (ne(B)->ne(arr(A,B))))).
% 1.94/2.13 all A B F (mem(F,arr(A,B))-> (all X (mem(X,A)->mem(ap(F,X),B)))).
% 1.94/2.13 all Q (mem(Q,bool)-> (all R (mem(R,bool)-> ((p(Q)<->p(R))->Q=R)))).
% 1.94/2.13 all A B F (mem(F,arr(A,B))-> (all G (mem(G,arr(A,B))-> ((all X (mem(X,A)->ap(F,X)=ap(G,X)))->F=G)))).
% 1.94/2.13 all A Y X (mem(X,A)->ap(k(A,Y),X)=Y).
% 1.94/2.13 all A X (mem(X,A)->ap(i(A),X)=X).
% 1.94/2.13 mem(c_2Ebool_2ET,bool).
% 1.94/2.13 p(c_2Ebool_2ET).
% 1.94/2.13 all A_27a (ne(A_27a)->mem(c_2Emin_2E_3D(A_27a),arr(A_27a,arr(A_27a,bool)))).
% 1.94/2.13 all A (ne(A)-> (all X (mem(X,A)-> (all Y (mem(Y,A)-> (p(ap(ap(c_2Emin_2E_3D(A),X),Y))<->X=Y)))))).
% 1.94/2.13 $T<->i(bool)=i(bool).
% 1.94/2.13 -$T.
% 1.94/2.13 end_of_list.
% 1.94/2.13
% 1.94/2.13 -------> usable clausifies to:
% 1.94/2.13
% 1.94/2.13 list(usable).
% 1.94/2.13 0 [] A=A.
% 1.94/2.13 0 [] ne(bool).
% 1.94/2.13 0 [] ne(ind).
% 1.94/2.13 0 [] -ne(A)| -ne(B)|ne(arr(A,B)).
% 1.94/2.13 0 [] -mem(F,arr(A,B))| -mem(X,A)|mem(ap(F,X),B).
% 1.94/2.13 0 [] -mem(Q,bool)| -mem(R,bool)|p(Q)|p(R)|Q=R.
% 1.94/2.13 0 [] -mem(Q,bool)| -mem(R,bool)| -p(Q)| -p(R)|Q=R.
% 1.94/2.13 0 [] -mem(F,arr(A,B))| -mem(G,arr(A,B))|mem($f1(A,B,F,G),A)|F=G.
% 1.94/2.13 0 [] -mem(F,arr(A,B))| -mem(G,arr(A,B))|ap(F,$f1(A,B,F,G))!=ap(G,$f1(A,B,F,G))|F=G.
% 1.94/2.13 0 [] -mem(X,A)|ap(k(A,Y),X)=Y.
% 1.94/2.13 0 [] -mem(X,A)|ap(i(A),X)=X.
% 1.94/2.13 0 [] mem(c_2Ebool_2ET,bool).
% 1.94/2.13 0 [] p(c_2Ebool_2ET).
% 1.94/2.13 0 [] -ne(A_27a)|mem(c_2Emin_2E_3D(A_27a),arr(A_27a,arr(A_27a,bool))).
% 1.94/2.13 0 [] -ne(A)| -mem(X,A)| -mem(Y,A)| -p(ap(ap(c_2Emin_2E_3D(A),X),Y))|X=Y.
% 1.94/2.13 0 [] -ne(A)| -mem(X,A)| -mem(Y,A)|p(ap(ap(c_2Emin_2E_3D(A),X),Y))|X!=Y.
% 1.94/2.13 0 [] -$T|i(bool)=i(bool).
% 1.94/2.13 0 [] $T|i(bool)!=i(bool).
% 1.94/2.13 0 [] -$T.
% 1.94/2.13 end_of_list.
% 1.94/2.13
% 1.94/2.13 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=5.
% 1.94/2.13
% 1.94/2.13 This ia a non-Horn set with equality. The strategy will be
% 1.94/2.13 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.94/2.13 deletion, with positive clauses in sos and nonpositive
% 1.94/2.13 clauses in usable.
% 1.94/2.13
% 1.94/2.13 dependent: set(knuth_bendix).
% 1.94/2.13 dependent: set(anl_eq).
% 1.94/2.13 dependent: set(para_from).
% 1.94/2.13 dependent: set(para_into).
% 1.94/2.13 dependent: clear(para_from_right).
% 1.94/2.13 dependent: clear(para_into_right).
% 1.94/2.13 dependent: set(para_from_vars).
% 1.94/2.13 dependent: set(eq_units_both_ways).
% 1.94/2.13 dependent: set(dynamic_demod_all).
% 1.94/2.13 dependent: set(dynamic_demod).
% 1.94/2.13 dependent: set(order_eq).
% 1.94/2.13 dependent: set(back_demod).
% 1.94/2.13 dependent: set(lrpo).
% 1.94/2.13 dependent: set(hyper_res).
% 1.94/2.13 dependent: set(unit_deletion).
% 1.94/2.13 dependent: set(factor).
% 1.94/2.13
% 1.94/2.13 ------------> process usable:
% 1.94/2.13 ** KEPT (pick-wt=8): 1 [] -ne(A)| -ne(B)|ne(arr(A,B)).
% 1.94/2.13 ** KEPT (pick-wt=13): 2 [] -mem(A,arr(B,C))| -mem(D,B)|mem(ap(A,D),C).
% 1.94/2.13 ** KEPT (pick-wt=13): 3 [] -mem(A,bool)| -mem(B,bool)|p(A)|p(B)|A=B.
% 1.94/2.13 ** KEPT (pick-wt=13): 4 [] -mem(A,bool)| -mem(B,bool)| -p(A)| -p(B)|A=B.
% 1.94/2.13 ** KEPT (pick-wt=20): 5 [] -mem(A,arr(B,C))| -mem(D,arr(B,C))|mem($f1(B,C,A,D),B)|A=D.
% 1.94/2.13 ** KEPT (pick-wt=28): 6 [] -mem(A,arr(B,C))| -mem(D,arr(B,C))|ap(A,$f1(B,C,A,D))!=ap(D,$f1(B,C,A,D))|A=D.
% 1.94/2.13 ** KEPT (pick-wt=10): 7 [] -mem(A,B)|ap(k(B,C),A)=C.
% 1.94/2.13 ** KEPT (pick-wt=9): 8 [] -mem(A,B)|ap(i(B),A)=A.
% 1.94/2.13 ** KEPT (pick-wt=10): 9 [] -ne(A)|mem(c_2Emin_2E_3D(A),arr(A,arr(A,bool))).
% 1.94/2.13 ** KEPT (pick-wt=18): 10 [] -ne(A)| -mem(B,A)| -mem(C,A)| -p(ap(ap(c_2Emin_2E_3D(A),B),C))|B=C.
% 1.94/2.13 ** KEPT (pick-wt=18): 11 [] -ne(A)| -mem(B,A)| -mem(C,A)|p(ap(ap(c_2Emin_2E_3D(A),B),C))|B!=C.
% 1.94/2.13 ** KEPT (pick-wt=5): 13 [copy,12,propositional] i(bool)=i(bool).
% 1.94/2.13 ** KEPT (pick-wt=0): 15 [copy,14,propositional] $F.
% 1.94/2.13
% 1.94/2.13 -----> EMPTY CLAUSE at 0.00 sec ----> 15 [copy,14,propositional] $F.
% 1.94/2.13
% 1.94/2.13 Length of proof is 0. Level of proof is 0.
% 1.94/2.13
% 1.94/2.13 ---------------- PROOF ----------------
% 1.94/2.13 % SZS status Theorem
% 1.94/2.13 % SZS output start Refutation
% See solution above
% 1.94/2.13 ------------ end of proof -------------
% 1.94/2.13
% 1.94/2.13
% 1.94/2.13 Search stopped by max_proofs option.
% 1.94/2.13
% 1.94/2.13
% 1.94/2.13 Search stopped by max_proofs option.
% 1.94/2.13
% 1.94/2.13 ============ end of search ============
% 1.94/2.13
% 1.94/2.13 -------------- statistics -------------
% 1.94/2.13 clauses given 0
% 1.94/2.13 clauses generated 0
% 1.94/2.13 clauses kept 12
% 1.94/2.13 clauses forward subsumed 0
% 1.94/2.13 clauses back subsumed 0
% 1.94/2.13 Kbytes malloced 976
% 1.94/2.13
% 1.94/2.13 ----------- times (seconds) -----------
% 1.94/2.13 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.94/2.13 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.94/2.13 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.94/2.13
% 1.94/2.13 That finishes the proof of the theorem.
% 1.94/2.13
% 1.94/2.13 Process 24116 finished Wed Jul 27 02:52:17 2022
% 1.94/2.13 Otter interrupted
% 1.94/2.13 PROOF FOUND
%------------------------------------------------------------------------------