TSTP Solution File: BOO001-1 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : BOO001-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n028.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 12:47:30 EDT 2022
% Result : Unsatisfiable 1.68s 1.86s
% Output : Refutation 1.68s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 5
% Syntax : Number of clauses : 14 ( 14 unt; 0 nHn; 2 RR)
% Number of literals : 14 ( 13 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 4 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 1 con; 0-3 aty)
% Number of variables : 35 ( 4 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
inverse(inverse(a)) != a,
file('BOO001-1.p',unknown),
[] ).
cnf(3,axiom,
multiply(multiply(A,B,C),D,multiply(A,B,E)) = multiply(A,B,multiply(C,D,E)),
file('BOO001-1.p',unknown),
[] ).
cnf(5,axiom,
multiply(A,B,B) = B,
file('BOO001-1.p',unknown),
[] ).
cnf(7,axiom,
multiply(A,A,B) = A,
file('BOO001-1.p',unknown),
[] ).
cnf(11,axiom,
multiply(A,B,inverse(B)) = A,
file('BOO001-1.p',unknown),
[] ).
cnf(13,plain,
multiply(A,B,multiply(A,C,D)) = multiply(A,C,multiply(inverse(C),B,D)),
inference(para_into,[status(thm),theory(equality)],[3,11]),
[iquote('para_into,3.1.1.1,11.1.1')] ).
cnf(16,plain,
multiply(A,B,multiply(C,A,D)) = multiply(C,A,multiply(A,B,D)),
inference(para_into,[status(thm),theory(equality)],[3,5]),
[iquote('para_into,3.1.1.1,5.1.1')] ).
cnf(39,plain,
multiply(A,B,C) = multiply(C,A,multiply(A,B,inverse(A))),
inference(para_into,[status(thm),theory(equality)],[16,11]),
[iquote('para_into,16.1.1.3,11.1.1')] ).
cnf(57,plain,
multiply(multiply(A,B,C),B,multiply(B,D,inverse(B))) = multiply(A,B,multiply(B,D,C)),
inference(flip,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[39,16])]),
[iquote('para_into,39.1.1,16.1.1,flip.1')] ).
cnf(80,plain,
multiply(A,B,multiply(inverse(B),A,C)) = A,
inference(flip,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[13,7])]),
[iquote('para_into,13.1.1,7.1.1,flip.1')] ).
cnf(126,plain,
multiply(inverse(A),B,multiply(B,A,C)) = B,
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[80,39]),57]),
[iquote('para_into,80.1.1,39.1.1,demod,57')] ).
cnf(242,plain,
multiply(inverse(A),B,A) = B,
inference(para_into,[status(thm),theory(equality)],[126,5]),
[iquote('para_into,126.1.1.3,5.1.1')] ).
cnf(264,plain,
inverse(inverse(A)) = A,
inference(para_into,[status(thm),theory(equality)],[242,11]),
[iquote('para_into,242.1.1,11.1.1')] ).
cnf(266,plain,
$false,
inference(binary,[status(thm)],[264,1]),
[iquote('binary,264.1,1.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.09/0.11 % Problem : BOO001-1 : TPTP v8.1.0. Released v1.0.0.
% 0.09/0.11 % Command : otter-tptp-script %s
% 0.11/0.32 % Computer : n028.cluster.edu
% 0.11/0.32 % Model : x86_64 x86_64
% 0.11/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.32 % Memory : 8042.1875MB
% 0.11/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.32 % CPULimit : 300
% 0.11/0.32 % WCLimit : 300
% 0.11/0.32 % DateTime : Wed Jul 27 02:40:03 EDT 2022
% 0.11/0.32 % CPUTime :
% 1.68/1.86 ----- Otter 3.3f, August 2004 -----
% 1.68/1.86 The process was started by sandbox2 on n028.cluster.edu,
% 1.68/1.86 Wed Jul 27 02:40:03 2022
% 1.68/1.86 The command was "./otter". The process ID is 4017.
% 1.68/1.86
% 1.68/1.86 set(prolog_style_variables).
% 1.68/1.86 set(auto).
% 1.68/1.86 dependent: set(auto1).
% 1.68/1.86 dependent: set(process_input).
% 1.68/1.86 dependent: clear(print_kept).
% 1.68/1.86 dependent: clear(print_new_demod).
% 1.68/1.86 dependent: clear(print_back_demod).
% 1.68/1.86 dependent: clear(print_back_sub).
% 1.68/1.86 dependent: set(control_memory).
% 1.68/1.86 dependent: assign(max_mem, 12000).
% 1.68/1.86 dependent: assign(pick_given_ratio, 4).
% 1.68/1.86 dependent: assign(stats_level, 1).
% 1.68/1.86 dependent: assign(max_seconds, 10800).
% 1.68/1.86 clear(print_given).
% 1.68/1.86
% 1.68/1.86 list(usable).
% 1.68/1.86 0 [] A=A.
% 1.68/1.86 0 [] multiply(multiply(V,W,X),Y,multiply(V,W,Z))=multiply(V,W,multiply(X,Y,Z)).
% 1.68/1.86 0 [] multiply(Y,X,X)=X.
% 1.68/1.86 0 [] multiply(X,X,Y)=X.
% 1.68/1.86 0 [] multiply(inverse(Y),Y,X)=X.
% 1.68/1.86 0 [] multiply(X,Y,inverse(Y))=X.
% 1.68/1.86 0 [] inverse(inverse(a))!=a.
% 1.68/1.86 end_of_list.
% 1.68/1.86
% 1.68/1.86 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.68/1.86
% 1.68/1.86 All clauses are units, and equality is present; the
% 1.68/1.86 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.68/1.86
% 1.68/1.86 dependent: set(knuth_bendix).
% 1.68/1.86 dependent: set(anl_eq).
% 1.68/1.86 dependent: set(para_from).
% 1.68/1.86 dependent: set(para_into).
% 1.68/1.86 dependent: clear(para_from_right).
% 1.68/1.86 dependent: clear(para_into_right).
% 1.68/1.86 dependent: set(para_from_vars).
% 1.68/1.86 dependent: set(eq_units_both_ways).
% 1.68/1.86 dependent: set(dynamic_demod_all).
% 1.68/1.86 dependent: set(dynamic_demod).
% 1.68/1.86 dependent: set(order_eq).
% 1.68/1.86 dependent: set(back_demod).
% 1.68/1.86 dependent: set(lrpo).
% 1.68/1.86
% 1.68/1.86 ------------> process usable:
% 1.68/1.86 ** KEPT (pick-wt=5): 1 [] inverse(inverse(a))!=a.
% 1.68/1.86
% 1.68/1.86 ------------> process sos:
% 1.68/1.86 ** KEPT (pick-wt=3): 2 [] A=A.
% 1.68/1.86 ** KEPT (pick-wt=18): 3 [] multiply(multiply(A,B,C),D,multiply(A,B,E))=multiply(A,B,multiply(C,D,E)).
% 1.68/1.86 ---> New Demodulator: 4 [new_demod,3] multiply(multiply(A,B,C),D,multiply(A,B,E))=multiply(A,B,multiply(C,D,E)).
% 1.68/1.86 ** KEPT (pick-wt=6): 5 [] multiply(A,B,B)=B.
% 1.68/1.86 ---> New Demodulator: 6 [new_demod,5] multiply(A,B,B)=B.
% 1.68/1.86 ** KEPT (pick-wt=6): 7 [] multiply(A,A,B)=A.
% 1.68/1.86 ---> New Demodulator: 8 [new_demod,7] multiply(A,A,B)=A.
% 1.68/1.86 ** KEPT (pick-wt=7): 9 [] multiply(inverse(A),A,B)=B.
% 1.68/1.86 ---> New Demodulator: 10 [new_demod,9] multiply(inverse(A),A,B)=B.
% 1.68/1.86 ** KEPT (pick-wt=7): 11 [] multiply(A,B,inverse(B))=A.
% 1.68/1.86 ---> New Demodulator: 12 [new_demod,11] multiply(A,B,inverse(B))=A.
% 1.68/1.86 Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.68/1.86 >>>> Starting back demodulation with 4.
% 1.68/1.86 >>>> Starting back demodulation with 6.
% 1.68/1.86 >>>> Starting back demodulation with 8.
% 1.68/1.86 >>>> Starting back demodulation with 10.
% 1.68/1.86 >>>> Starting back demodulation with 12.
% 1.68/1.86
% 1.68/1.86 ======= end of input processing =======
% 1.68/1.86
% 1.68/1.86 =========== start of search ===========
% 1.68/1.86
% 1.68/1.86 -------- PROOF --------
% 1.68/1.86
% 1.68/1.86 ----> UNIT CONFLICT at 0.01 sec ----> 266 [binary,264.1,1.1] $F.
% 1.68/1.86
% 1.68/1.86 Length of proof is 8. Level of proof is 6.
% 1.68/1.86
% 1.68/1.86 ---------------- PROOF ----------------
% 1.68/1.86 % SZS status Unsatisfiable
% 1.68/1.86 % SZS output start Refutation
% See solution above
% 1.68/1.86 ------------ end of proof -------------
% 1.68/1.86
% 1.68/1.86
% 1.68/1.86 Search stopped by max_proofs option.
% 1.68/1.86
% 1.68/1.86
% 1.68/1.86 Search stopped by max_proofs option.
% 1.68/1.86
% 1.68/1.86 ============ end of search ============
% 1.68/1.86
% 1.68/1.86 -------------- statistics -------------
% 1.68/1.86 clauses given 18
% 1.68/1.86 clauses generated 379
% 1.68/1.86 clauses kept 165
% 1.68/1.86 clauses forward subsumed 308
% 1.68/1.86 clauses back subsumed 0
% 1.68/1.86 Kbytes malloced 1953
% 1.68/1.86
% 1.68/1.86 ----------- times (seconds) -----------
% 1.68/1.86 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.68/1.86 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.86 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.68/1.86
% 1.68/1.86 That finishes the proof of the theorem.
% 1.68/1.86
% 1.68/1.86 Process 4017 finished Wed Jul 27 02:40:05 2022
% 1.68/1.86 Otter interrupted
% 1.68/1.86 PROOF FOUND
%------------------------------------------------------------------------------