TSTP Solution File: COL032-1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : COL032-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n003.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 12:48:02 EDT 2022
% Result : Unsatisfiable 1.66s 1.88s
% Output : Refutation 1.66s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 3
% Syntax : Number of clauses : 9 ( 9 unt; 0 nHn; 4 RR)
% Number of literals : 9 ( 8 equ; 3 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 7 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 2 con; 0-2 aty)
% Number of variables : 13 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
apply(A,f(A)) != apply(f(A),apply(A,f(A))),
file('COL032-1.p',unknown),
[] ).
cnf(2,plain,
apply(f(A),apply(A,f(A))) != apply(A,f(A)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
[iquote('copy,1,flip.1')] ).
cnf(4,axiom,
apply(m,A) = apply(A,A),
file('COL032-1.p',unknown),
[] ).
cnf(6,axiom,
apply(apply(apply(q,A),B),C) = apply(B,apply(A,C)),
file('COL032-1.p',unknown),
[] ).
cnf(7,plain,
apply(A,A) = apply(m,A),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[4])]),
[iquote('copy,4,flip.1')] ).
cnf(13,plain,
apply(m,apply(apply(q,A),B)) = apply(B,apply(A,apply(apply(q,A),B))),
inference(para_into,[status(thm),theory(equality)],[6,7]),
[iquote('para_into,5.1.1,7.1.1')] ).
cnf(14,plain,
apply(A,apply(B,apply(apply(q,B),A))) = apply(m,apply(apply(q,B),A)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[13])]),
[iquote('copy,13,flip.1')] ).
cnf(15,plain,
apply(f(apply(apply(q,A),B)),apply(B,apply(A,f(apply(apply(q,A),B))))) != apply(B,apply(A,f(apply(apply(q,A),B)))),
inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[6,2]),6]),
[iquote('para_from,5.1.1,2.1.1.2,demod,6')] ).
cnf(16,plain,
$false,
inference(binary,[status(thm)],[15,14]),
[iquote('binary,15.1,14.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : COL032-1 : TPTP v8.1.0. Released v1.0.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n003.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Wed Jul 27 02:23:56 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.66/1.88 ----- Otter 3.3f, August 2004 -----
% 1.66/1.88 The process was started by sandbox on n003.cluster.edu,
% 1.66/1.88 Wed Jul 27 02:23:56 2022
% 1.66/1.88 The command was "./otter". The process ID is 21446.
% 1.66/1.88
% 1.66/1.88 set(prolog_style_variables).
% 1.66/1.88 set(auto).
% 1.66/1.88 dependent: set(auto1).
% 1.66/1.88 dependent: set(process_input).
% 1.66/1.88 dependent: clear(print_kept).
% 1.66/1.88 dependent: clear(print_new_demod).
% 1.66/1.88 dependent: clear(print_back_demod).
% 1.66/1.88 dependent: clear(print_back_sub).
% 1.66/1.88 dependent: set(control_memory).
% 1.66/1.88 dependent: assign(max_mem, 12000).
% 1.66/1.88 dependent: assign(pick_given_ratio, 4).
% 1.66/1.88 dependent: assign(stats_level, 1).
% 1.66/1.88 dependent: assign(max_seconds, 10800).
% 1.66/1.88 clear(print_given).
% 1.66/1.88
% 1.66/1.88 list(usable).
% 1.66/1.88 0 [] A=A.
% 1.66/1.88 0 [] apply(m,X)=apply(X,X).
% 1.66/1.88 0 [] apply(apply(apply(q,X),Y),Z)=apply(Y,apply(X,Z)).
% 1.66/1.88 0 [] apply(Y,f(Y))!=apply(f(Y),apply(Y,f(Y))).
% 1.66/1.88 end_of_list.
% 1.66/1.88
% 1.66/1.88 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.66/1.88
% 1.66/1.88 All clauses are units, and equality is present; the
% 1.66/1.88 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.66/1.88
% 1.66/1.88 dependent: set(knuth_bendix).
% 1.66/1.88 dependent: set(anl_eq).
% 1.66/1.88 dependent: set(para_from).
% 1.66/1.88 dependent: set(para_into).
% 1.66/1.88 dependent: clear(para_from_right).
% 1.66/1.88 dependent: clear(para_into_right).
% 1.66/1.88 dependent: set(para_from_vars).
% 1.66/1.88 dependent: set(eq_units_both_ways).
% 1.66/1.88 dependent: set(dynamic_demod_all).
% 1.66/1.88 dependent: set(dynamic_demod).
% 1.66/1.88 dependent: set(order_eq).
% 1.66/1.88 dependent: set(back_demod).
% 1.66/1.88 dependent: set(lrpo).
% 1.66/1.88
% 1.66/1.88 ------------> process usable:
% 1.66/1.88 ** KEPT (pick-wt=12): 2 [copy,1,flip.1] apply(f(A),apply(A,f(A)))!=apply(A,f(A)).
% 1.66/1.88
% 1.66/1.88 ------------> process sos:
% 1.66/1.88 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.66/1.88 ** KEPT (pick-wt=7): 4 [] apply(m,A)=apply(A,A).
% 1.66/1.88 ** KEPT (pick-wt=13): 5 [] apply(apply(apply(q,A),B),C)=apply(B,apply(A,C)).
% 1.66/1.88 ---> New Demodulator: 6 [new_demod,5] apply(apply(apply(q,A),B),C)=apply(B,apply(A,C)).
% 1.66/1.88 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.66/1.88 ** KEPT (pick-wt=7): 7 [copy,4,flip.1] apply(A,A)=apply(m,A).
% 1.66/1.88 >>>> Starting back demodulation with 6.
% 1.66/1.88 Following clause subsumed by 4 during input processing: 0 [copy,7,flip.1] apply(m,A)=apply(A,A).
% 1.66/1.88
% 1.66/1.88 ======= end of input processing =======
% 1.66/1.88
% 1.66/1.88 =========== start of search ===========
% 1.66/1.88
% 1.66/1.88 -------- PROOF --------
% 1.66/1.88
% 1.66/1.88 ----> UNIT CONFLICT at 0.00 sec ----> 16 [binary,15.1,14.1] $F.
% 1.66/1.88
% 1.66/1.88 Length of proof is 5. Level of proof is 3.
% 1.66/1.88
% 1.66/1.88 ---------------- PROOF ----------------
% 1.66/1.88 % SZS status Unsatisfiable
% 1.66/1.88 % SZS output start Refutation
% See solution above
% 1.66/1.88 ------------ end of proof -------------
% 1.66/1.88
% 1.66/1.88
% 1.66/1.88 Search stopped by max_proofs option.
% 1.66/1.88
% 1.66/1.88
% 1.66/1.88 Search stopped by max_proofs option.
% 1.66/1.88
% 1.66/1.88 ============ end of search ============
% 1.66/1.88
% 1.66/1.88 -------------- statistics -------------
% 1.66/1.88 clauses given 4
% 1.66/1.88 clauses generated 14
% 1.66/1.88 clauses kept 11
% 1.66/1.88 clauses forward subsumed 12
% 1.66/1.88 clauses back subsumed 0
% 1.66/1.88 Kbytes malloced 976
% 1.66/1.88
% 1.66/1.88 ----------- times (seconds) -----------
% 1.66/1.88 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.88 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.88 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.66/1.88
% 1.66/1.88 That finishes the proof of the theorem.
% 1.66/1.88
% 1.66/1.88 Process 21446 finished Wed Jul 27 02:23:58 2022
% 1.66/1.88 Otter interrupted
% 1.66/1.88 PROOF FOUND
%------------------------------------------------------------------------------