TSTP Solution File: GRP012-4 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : GRP012-4 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n020.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 12:55:51 EDT 2022
% Result : Unsatisfiable 1.90s 2.10s
% Output : Refutation 1.90s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 6
% Syntax : Number of clauses : 11 ( 11 unt; 0 nHn; 2 RR)
% Number of literals : 11 ( 10 equ; 1 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 5 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 3 con; 0-2 aty)
% Number of variables : 15 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
inverse(multiply(a,b)) != multiply(inverse(b),inverse(a)),
file('GRP012-4.p',unknown),
[] ).
cnf(4,axiom,
multiply(identity,A) = A,
file('GRP012-4.p',unknown),
[] ).
cnf(5,axiom,
multiply(inverse(A),A) = identity,
file('GRP012-4.p',unknown),
[] ).
cnf(7,axiom,
multiply(multiply(A,B),C) = multiply(A,multiply(B,C)),
file('GRP012-4.p',unknown),
[] ).
cnf(10,axiom,
multiply(A,identity) = A,
file('GRP012-4.p',unknown),
[] ).
cnf(11,axiom,
multiply(A,inverse(A)) = identity,
file('GRP012-4.p',unknown),
[] ).
cnf(15,plain,
multiply(inverse(A),multiply(A,B)) = B,
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[7,5]),4])]),
[iquote('para_into,7.1.1.1,5.1.1,demod,4,flip.1')] ).
cnf(17,plain,
multiply(A,multiply(B,inverse(multiply(A,B)))) = identity,
inference(para_into,[status(thm),theory(equality)],[11,7]),
[iquote('para_into,11.1.1,7.1.1')] ).
cnf(33,plain,
multiply(A,inverse(multiply(B,A))) = inverse(B),
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[17,15]),10])]),
[iquote('para_from,17.1.1,15.1.1.2,demod,10,flip.1')] ).
cnf(41,plain,
inverse(multiply(A,B)) = multiply(inverse(B),inverse(A)),
inference(flip,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[33,15])]),
[iquote('para_from,33.1.1,15.1.1.2,flip.1')] ).
cnf(43,plain,
$false,
inference(binary,[status(thm)],[41,1]),
[iquote('binary,41.1,1.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : GRP012-4 : TPTP v8.1.0. Released v1.0.0.
% 0.00/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n020.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 05:37:38 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.90/2.10 ----- Otter 3.3f, August 2004 -----
% 1.90/2.10 The process was started by sandbox2 on n020.cluster.edu,
% 1.90/2.10 Wed Jul 27 05:37:38 2022
% 1.90/2.10 The command was "./otter". The process ID is 5482.
% 1.90/2.10
% 1.90/2.10 set(prolog_style_variables).
% 1.90/2.10 set(auto).
% 1.90/2.10 dependent: set(auto1).
% 1.90/2.10 dependent: set(process_input).
% 1.90/2.10 dependent: clear(print_kept).
% 1.90/2.10 dependent: clear(print_new_demod).
% 1.90/2.10 dependent: clear(print_back_demod).
% 1.90/2.10 dependent: clear(print_back_sub).
% 1.90/2.10 dependent: set(control_memory).
% 1.90/2.10 dependent: assign(max_mem, 12000).
% 1.90/2.10 dependent: assign(pick_given_ratio, 4).
% 1.90/2.10 dependent: assign(stats_level, 1).
% 1.90/2.10 dependent: assign(max_seconds, 10800).
% 1.90/2.10 clear(print_given).
% 1.90/2.10
% 1.90/2.10 list(usable).
% 1.90/2.10 0 [] A=A.
% 1.90/2.10 0 [] multiply(identity,X)=X.
% 1.90/2.10 0 [] multiply(inverse(X),X)=identity.
% 1.90/2.10 0 [] multiply(multiply(X,Y),Z)=multiply(X,multiply(Y,Z)).
% 1.90/2.10 0 [] multiply(X,identity)=X.
% 1.90/2.10 0 [] multiply(X,inverse(X))=identity.
% 1.90/2.10 0 [] inverse(multiply(a,b))!=multiply(inverse(b),inverse(a)).
% 1.90/2.10 end_of_list.
% 1.90/2.10
% 1.90/2.10 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.90/2.10
% 1.90/2.10 All clauses are units, and equality is present; the
% 1.90/2.10 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.90/2.10
% 1.90/2.10 dependent: set(knuth_bendix).
% 1.90/2.10 dependent: set(anl_eq).
% 1.90/2.10 dependent: set(para_from).
% 1.90/2.10 dependent: set(para_into).
% 1.90/2.10 dependent: clear(para_from_right).
% 1.90/2.10 dependent: clear(para_into_right).
% 1.90/2.10 dependent: set(para_from_vars).
% 1.90/2.10 dependent: set(eq_units_both_ways).
% 1.90/2.10 dependent: set(dynamic_demod_all).
% 1.90/2.10 dependent: set(dynamic_demod).
% 1.90/2.10 dependent: set(order_eq).
% 1.90/2.10 dependent: set(back_demod).
% 1.90/2.10 dependent: set(lrpo).
% 1.90/2.10
% 1.90/2.10 ------------> process usable:
% 1.90/2.10 ** KEPT (pick-wt=10): 1 [] inverse(multiply(a,b))!=multiply(inverse(b),inverse(a)).
% 1.90/2.10
% 1.90/2.10 ------------> process sos:
% 1.90/2.10 ** KEPT (pick-wt=3): 2 [] A=A.
% 1.90/2.10 ** KEPT (pick-wt=5): 3 [] multiply(identity,A)=A.
% 1.90/2.10 ---> New Demodulator: 4 [new_demod,3] multiply(identity,A)=A.
% 1.90/2.10 ** KEPT (pick-wt=6): 5 [] multiply(inverse(A),A)=identity.
% 1.90/2.10 ---> New Demodulator: 6 [new_demod,5] multiply(inverse(A),A)=identity.
% 1.90/2.10 ** KEPT (pick-wt=11): 7 [] multiply(multiply(A,B),C)=multiply(A,multiply(B,C)).
% 1.90/2.10 ---> New Demodulator: 8 [new_demod,7] multiply(multiply(A,B),C)=multiply(A,multiply(B,C)).
% 1.90/2.10 ** KEPT (pick-wt=5): 9 [] multiply(A,identity)=A.
% 1.90/2.10 ---> New Demodulator: 10 [new_demod,9] multiply(A,identity)=A.
% 1.90/2.10 ** KEPT (pick-wt=6): 11 [] multiply(A,inverse(A))=identity.
% 1.90/2.10 ---> New Demodulator: 12 [new_demod,11] multiply(A,inverse(A))=identity.
% 1.90/2.10 Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.90/2.10 >>>> Starting back demodulation with 4.
% 1.90/2.10 >>>> Starting back demodulation with 6.
% 1.90/2.10 >>>> Starting back demodulation with 8.
% 1.90/2.10 >>>> Starting back demodulation with 10.
% 1.90/2.10 >>>> Starting back demodulation with 12.
% 1.90/2.10
% 1.90/2.10 ======= end of input processing =======
% 1.90/2.10
% 1.90/2.10 =========== start of search ===========
% 1.90/2.10
% 1.90/2.10 -------- PROOF --------
% 1.90/2.10
% 1.90/2.10 ----> UNIT CONFLICT at 0.00 sec ----> 43 [binary,41.1,1.1] $F.
% 1.90/2.10
% 1.90/2.10 Length of proof is 4. Level of proof is 3.
% 1.90/2.10
% 1.90/2.10 ---------------- PROOF ----------------
% 1.90/2.10 % SZS status Unsatisfiable
% 1.90/2.10 % SZS output start Refutation
% See solution above
% 1.90/2.10 ------------ end of proof -------------
% 1.90/2.10
% 1.90/2.10
% 1.90/2.10 Search stopped by max_proofs option.
% 1.90/2.10
% 1.90/2.10
% 1.90/2.10 Search stopped by max_proofs option.
% 1.90/2.10
% 1.90/2.10 ============ end of search ============
% 1.90/2.10
% 1.90/2.10 -------------- statistics -------------
% 1.90/2.10 clauses given 12
% 1.90/2.10 clauses generated 104
% 1.90/2.10 clauses kept 22
% 1.90/2.10 clauses forward subsumed 94
% 1.90/2.10 clauses back subsumed 0
% 1.90/2.10 Kbytes malloced 976
% 1.90/2.10
% 1.90/2.10 ----------- times (seconds) -----------
% 1.90/2.10 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.90/2.10 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.90/2.10 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.90/2.10
% 1.90/2.10 That finishes the proof of the theorem.
% 1.90/2.10
% 1.90/2.10 Process 5482 finished Wed Jul 27 05:37:40 2022
% 1.90/2.10 Otter interrupted
% 1.90/2.10 PROOF FOUND
%------------------------------------------------------------------------------