TSTP Solution File: GRP542-1 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : GRP542-1 : TPTP v8.1.0. Released v2.6.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n027.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 12:57:13 EDT 2022
% Result : Unsatisfiable 1.66s 1.87s
% Output : Refutation 1.66s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 5
% Syntax : Number of clauses : 10 ( 10 unt; 0 nHn; 3 RR)
% Number of literals : 10 ( 9 equ; 2 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 6 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 3 con; 0-2 aty)
% Number of variables : 11 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
multiply(multiply(inverse(b2),b2),a2) != a2,
file('GRP542-1.p',unknown),
[] ).
cnf(3,axiom,
divide(divide(identity,divide(divide(divide(A,B),C),A)),C) = B,
file('GRP542-1.p',unknown),
[] ).
cnf(6,axiom,
multiply(A,B) = divide(A,divide(identity,B)),
file('GRP542-1.p',unknown),
[] ).
cnf(8,axiom,
inverse(A) = divide(identity,A),
file('GRP542-1.p',unknown),
[] ).
cnf(9,axiom,
identity = divide(A,A),
file('GRP542-1.p',unknown),
[] ).
cnf(11,plain,
divide(A,A) = identity,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[9])]),
[iquote('copy,9,flip.1')] ).
cnf(12,plain,
divide(identity,divide(identity,a2)) != a2,
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[1]),8,6,11,6]),
[iquote('back_demod,1,demod,8,6,11,6')] ).
cnf(17,plain,
divide(divide(identity,divide(identity,A)),divide(A,B)) = B,
inference(para_into,[status(thm),theory(equality)],[3,11]),
[iquote('para_into,3.1.1.1.2.1,10.1.1')] ).
cnf(29,plain,
divide(identity,divide(identity,A)) = A,
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[17,11]),11]),
[iquote('para_into,17.1.1.1.2,10.1.1,demod,11')] ).
cnf(31,plain,
$false,
inference(binary,[status(thm)],[29,12]),
[iquote('binary,29.1,12.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.11 % Problem : GRP542-1 : TPTP v8.1.0. Released v2.6.0.
% 0.06/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n027.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 05:42:36 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.66/1.87 ----- Otter 3.3f, August 2004 -----
% 1.66/1.87 The process was started by sandbox on n027.cluster.edu,
% 1.66/1.87 Wed Jul 27 05:42:36 2022
% 1.66/1.87 The command was "./otter". The process ID is 16588.
% 1.66/1.87
% 1.66/1.87 set(prolog_style_variables).
% 1.66/1.87 set(auto).
% 1.66/1.87 dependent: set(auto1).
% 1.66/1.87 dependent: set(process_input).
% 1.66/1.87 dependent: clear(print_kept).
% 1.66/1.87 dependent: clear(print_new_demod).
% 1.66/1.87 dependent: clear(print_back_demod).
% 1.66/1.87 dependent: clear(print_back_sub).
% 1.66/1.87 dependent: set(control_memory).
% 1.66/1.87 dependent: assign(max_mem, 12000).
% 1.66/1.87 dependent: assign(pick_given_ratio, 4).
% 1.66/1.87 dependent: assign(stats_level, 1).
% 1.66/1.87 dependent: assign(max_seconds, 10800).
% 1.66/1.87 clear(print_given).
% 1.66/1.87
% 1.66/1.87 list(usable).
% 1.66/1.87 0 [] A=A.
% 1.66/1.87 0 [] divide(divide(identity,divide(divide(divide(A,B),C),A)),C)=B.
% 1.66/1.87 0 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.66/1.87 0 [] inverse(A)=divide(identity,A).
% 1.66/1.87 0 [] identity=divide(A,A).
% 1.66/1.87 0 [] multiply(multiply(inverse(b2),b2),a2)!=a2.
% 1.66/1.87 end_of_list.
% 1.66/1.87
% 1.66/1.87 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.66/1.87
% 1.66/1.87 All clauses are units, and equality is present; the
% 1.66/1.87 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.66/1.87
% 1.66/1.87 dependent: set(knuth_bendix).
% 1.66/1.87 dependent: set(anl_eq).
% 1.66/1.87 dependent: set(para_from).
% 1.66/1.87 dependent: set(para_into).
% 1.66/1.87 dependent: clear(para_from_right).
% 1.66/1.87 dependent: clear(para_into_right).
% 1.66/1.87 dependent: set(para_from_vars).
% 1.66/1.87 dependent: set(eq_units_both_ways).
% 1.66/1.87 dependent: set(dynamic_demod_all).
% 1.66/1.87 dependent: set(dynamic_demod).
% 1.66/1.87 dependent: set(order_eq).
% 1.66/1.87 dependent: set(back_demod).
% 1.66/1.87 dependent: set(lrpo).
% 1.66/1.87
% 1.66/1.87 ------------> process usable:
% 1.66/1.87 ** KEPT (pick-wt=8): 1 [] multiply(multiply(inverse(b2),b2),a2)!=a2.
% 1.66/1.87
% 1.66/1.87 ------------> process sos:
% 1.66/1.87 ** KEPT (pick-wt=3): 2 [] A=A.
% 1.66/1.87 ** KEPT (pick-wt=13): 3 [] divide(divide(identity,divide(divide(divide(A,B),C),A)),C)=B.
% 1.66/1.87 ---> New Demodulator: 4 [new_demod,3] divide(divide(identity,divide(divide(divide(A,B),C),A)),C)=B.
% 1.66/1.87 ** KEPT (pick-wt=9): 5 [] multiply(A,B)=divide(A,divide(identity,B)).
% 1.66/1.87 ---> New Demodulator: 6 [new_demod,5] multiply(A,B)=divide(A,divide(identity,B)).
% 1.66/1.87 ** KEPT (pick-wt=6): 7 [] inverse(A)=divide(identity,A).
% 1.66/1.87 ---> New Demodulator: 8 [new_demod,7] inverse(A)=divide(identity,A).
% 1.66/1.87 ** KEPT (pick-wt=5): 10 [copy,9,flip.1] divide(A,A)=identity.
% 1.66/1.87 ---> New Demodulator: 11 [new_demod,10] divide(A,A)=identity.
% 1.66/1.87 Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.66/1.87 >>>> Starting back demodulation with 4.
% 1.66/1.87 >>>> Starting back demodulation with 6.
% 1.66/1.87 >> back demodulating 1 with 6.
% 1.66/1.87 >>>> Starting back demodulation with 8.
% 1.66/1.87 >>>> Starting back demodulation with 11.
% 1.66/1.87
% 1.66/1.87 ======= end of input processing =======
% 1.66/1.87
% 1.66/1.87 =========== start of search ===========
% 1.66/1.87
% 1.66/1.87 -------- PROOF --------
% 1.66/1.87
% 1.66/1.87 ----> UNIT CONFLICT at 0.00 sec ----> 31 [binary,29.1,12.1] $F.
% 1.66/1.87
% 1.66/1.87 Length of proof is 4. Level of proof is 3.
% 1.66/1.87
% 1.66/1.87 ---------------- PROOF ----------------
% 1.66/1.87 % SZS status Unsatisfiable
% 1.66/1.87 % SZS output start Refutation
% See solution above
% 1.66/1.87 ------------ end of proof -------------
% 1.66/1.87
% 1.66/1.87
% 1.66/1.87 Search stopped by max_proofs option.
% 1.66/1.87
% 1.66/1.87
% 1.66/1.87 Search stopped by max_proofs option.
% 1.66/1.87
% 1.66/1.87 ============ end of search ============
% 1.66/1.87
% 1.66/1.87 -------------- statistics -------------
% 1.66/1.87 clauses given 9
% 1.66/1.87 clauses generated 22
% 1.66/1.87 clauses kept 16
% 1.66/1.87 clauses forward subsumed 16
% 1.66/1.87 clauses back subsumed 0
% 1.66/1.87 Kbytes malloced 976
% 1.66/1.87
% 1.66/1.87 ----------- times (seconds) -----------
% 1.66/1.87 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.87 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.87 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.66/1.87
% 1.66/1.87 That finishes the proof of the theorem.
% 1.66/1.87
% 1.66/1.87 Process 16588 finished Wed Jul 27 05:42:38 2022
% 1.66/1.87 Otter interrupted
% 1.66/1.87 PROOF FOUND
%------------------------------------------------------------------------------