TSTP Solution File: GRP450-1 by Prover9---1109a
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- Process Solution
%------------------------------------------------------------------------------
% File : Prover9---1109a
% Problem : GRP450-1 : TPTP v8.1.0. Released v2.6.0.
% Transfm : none
% Format : tptp:raw
% Command : tptp2X_and_run_prover9 %d %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 : 600s
% DateTime : Sat Jul 16 11:19:08 EDT 2022
% Result : Unsatisfiable 0.75s 1.02s
% Output : Refutation 0.75s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12 % Problem : GRP450-1 : TPTP v8.1.0. Released v2.6.0.
% 0.12/0.13 % Command : tptp2X_and_run_prover9 %d %s
% 0.13/0.34 % Computer : n020.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 : 600
% 0.13/0.34 % DateTime : Tue Jun 14 02:32:05 EDT 2022
% 0.13/0.35 % CPUTime :
% 0.75/1.02 ============================== Prover9 ===============================
% 0.75/1.02 Prover9 (32) version 2009-11A, November 2009.
% 0.75/1.02 Process 25635 was started by sandbox on n020.cluster.edu,
% 0.75/1.02 Tue Jun 14 02:32:05 2022
% 0.75/1.02 The command was "/export/starexec/sandbox/solver/bin/prover9 -t 300 -f /tmp/Prover9_25482_n020.cluster.edu".
% 0.75/1.02 ============================== end of head ===========================
% 0.75/1.02
% 0.75/1.02 ============================== INPUT =================================
% 0.75/1.02
% 0.75/1.02 % Reading from file /tmp/Prover9_25482_n020.cluster.edu
% 0.75/1.02
% 0.75/1.02 set(prolog_style_variables).
% 0.75/1.02 set(auto2).
% 0.75/1.02 % set(auto2) -> set(auto).
% 0.75/1.02 % set(auto) -> set(auto_inference).
% 0.75/1.02 % set(auto) -> set(auto_setup).
% 0.75/1.02 % set(auto_setup) -> set(predicate_elim).
% 0.75/1.02 % set(auto_setup) -> assign(eq_defs, unfold).
% 0.75/1.02 % set(auto) -> set(auto_limits).
% 0.75/1.02 % set(auto_limits) -> assign(max_weight, "100.000").
% 0.75/1.02 % set(auto_limits) -> assign(sos_limit, 20000).
% 0.75/1.02 % set(auto) -> set(auto_denials).
% 0.75/1.02 % set(auto) -> set(auto_process).
% 0.75/1.02 % set(auto2) -> assign(new_constants, 1).
% 0.75/1.02 % set(auto2) -> assign(fold_denial_max, 3).
% 0.75/1.02 % set(auto2) -> assign(max_weight, "200.000").
% 0.75/1.02 % set(auto2) -> assign(max_hours, 1).
% 0.75/1.02 % assign(max_hours, 1) -> assign(max_seconds, 3600).
% 0.75/1.02 % set(auto2) -> assign(max_seconds, 0).
% 0.75/1.02 % set(auto2) -> assign(max_minutes, 5).
% 0.75/1.02 % assign(max_minutes, 5) -> assign(max_seconds, 300).
% 0.75/1.02 % set(auto2) -> set(sort_initial_sos).
% 0.75/1.02 % set(auto2) -> assign(sos_limit, -1).
% 0.75/1.02 % set(auto2) -> assign(lrs_ticks, 3000).
% 0.75/1.02 % set(auto2) -> assign(max_megs, 400).
% 0.75/1.02 % set(auto2) -> assign(stats, some).
% 0.75/1.02 % set(auto2) -> clear(echo_input).
% 0.75/1.02 % set(auto2) -> set(quiet).
% 0.75/1.02 % set(auto2) -> clear(print_initial_clauses).
% 0.75/1.02 % set(auto2) -> clear(print_given).
% 0.75/1.02 assign(lrs_ticks,-1).
% 0.75/1.02 assign(sos_limit,10000).
% 0.75/1.02 assign(order,kbo).
% 0.75/1.02 set(lex_order_vars).
% 0.75/1.02 clear(print_given).
% 0.75/1.02
% 0.75/1.02 % formulas(sos). % not echoed (4 formulas)
% 0.75/1.02
% 0.75/1.02 ============================== end of input ==========================
% 0.75/1.02
% 0.75/1.02 % From the command line: assign(max_seconds, 300).
% 0.75/1.02
% 0.75/1.02 ============================== PROCESS NON-CLAUSAL FORMULAS ==========
% 0.75/1.02
% 0.75/1.02 % Formulas that are not ordinary clauses:
% 0.75/1.02
% 0.75/1.02 ============================== end of process non-clausal formulas ===
% 0.75/1.02
% 0.75/1.02 ============================== PROCESS INITIAL CLAUSES ===============
% 0.75/1.02
% 0.75/1.02 ============================== PREDICATE ELIMINATION =================
% 0.75/1.02
% 0.75/1.02 ============================== end predicate elimination =============
% 0.75/1.02
% 0.75/1.02 Auto_denials:
% 0.75/1.02 % copying label prove_these_axioms_3 to answer in negative clause
% 0.75/1.02
% 0.75/1.02 Term ordering decisions:
% 0.75/1.02
% 0.75/1.02 % Assigning unary symbol inverse kb_weight 0 and highest precedence (7).
% 0.75/1.02 Function symbol KB weights: a3=1. b3=1. c3=1. divide=1. multiply=1. inverse=0.
% 0.75/1.02
% 0.75/1.02 ============================== end of process initial clauses ========
% 0.75/1.02
% 0.75/1.02 ============================== CLAUSES FOR SEARCH ====================
% 0.75/1.02
% 0.75/1.02 ============================== end of clauses for search =============
% 0.75/1.02
% 0.75/1.02 ============================== SEARCH ================================
% 0.75/1.02
% 0.75/1.02 % Starting search at 0.01 seconds.
% 0.75/1.02
% 0.75/1.02 ============================== PROOF =================================
% 0.75/1.02 % SZS status Unsatisfiable
% 0.75/1.02 % SZS output start Refutation
% 0.75/1.02
% 0.75/1.02 % Proof 1 at 0.02 (+ 0.00) seconds: prove_these_axioms_3.
% 0.75/1.02 % Length of proof is 34.
% 0.75/1.02 % Level of proof is 16.
% 0.75/1.02 % Maximum clause weight is 16.000.
% 0.75/1.02 % Given clauses 27.
% 0.75/1.02
% 0.75/1.02 1 inverse(A) = divide(divide(B,B),A) # label(inverse) # label(axiom). [assumption].
% 0.75/1.02 2 divide(divide(A,A),B) = inverse(B). [copy(1),flip(a)].
% 0.75/1.02 3 multiply(A,B) = divide(A,divide(divide(C,C),B)) # label(multiply) # label(axiom). [assumption].
% 0.75/1.02 4 multiply(A,B) = divide(A,inverse(B)). [copy(3),rewrite([2(3)])].
% 0.75/1.02 5 divide(A,divide(divide(divide(divide(B,B),B),C),divide(divide(divide(B,B),A),C))) = B # label(single_axiom) # label(axiom). [assumption].
% 0.75/1.02 6 divide(A,divide(divide(inverse(B),C),divide(inverse(A),C))) = B. [copy(5),rewrite([2(2),2(4)])].
% 0.75/1.02 7 multiply(multiply(a3,b3),c3) != multiply(a3,multiply(b3,c3)) # label(prove_these_axioms_3) # label(negated_conjecture) # answer(prove_these_axioms_3). [assumption].
% 0.75/1.02 8 divide(divide(a3,inverse(b3)),inverse(c3)) != divide(a3,inverse(divide(b3,inverse(c3)))) # answer(prove_these_axioms_3). [copy(7),rewrite([4(3),4(6),4(11),4(13)])].
% 0.75/1.02 9 divide(inverse(divide(A,A)),B) = inverse(B). [para(2(a,1),2(a,1,1))].
% 0.75/1.02 10 inverse(divide(divide(inverse(A),B),inverse(B))) = A. [para(6(a,1),2(a,1)),rewrite([9(5)]),flip(a)].
% 0.75/1.02 20 inverse(divide(A,inverse(divide(divide(inverse(A),B),divide(inverse(inverse(C)),B))))) = C. [para(6(a,1),10(a,1,1,1))].
% 0.75/1.02 21 inverse(divide(inverse(A),inverse(A))) = divide(B,B). [para(9(a,1),10(a,1,1,1))].
% 0.75/1.02 24 inverse(divide(inverse(A),inverse(A))) = c_0. [new_symbol(21)].
% 0.75/1.02 25 divide(A,A) = c_0. [back_rewrite(21),rewrite([24(4)]),flip(a)].
% 0.75/1.02 26 inverse(c_0) = c_0. [back_rewrite(24),rewrite([25(3)])].
% 0.75/1.02 29 inverse(divide(divide(inverse(A),c_0),c_0)) = A. [para(26(a,1),10(a,1,1,2))].
% 0.75/1.02 30 divide(A,divide(divide(inverse(B),inverse(A)),c_0)) = B. [para(25(a,1),6(a,1,2,2))].
% 0.75/1.02 31 divide(A,c_0) = A. [para(25(a,1),6(a,1,2))].
% 0.75/1.02 32 divide(A,divide(inverse(B),inverse(A))) = B. [back_rewrite(30),rewrite([31(5)])].
% 0.75/1.02 33 inverse(inverse(A)) = A. [back_rewrite(29),rewrite([31(3),31(3)])].
% 0.75/1.02 34 inverse(divide(A,inverse(divide(divide(inverse(A),B),divide(C,B))))) = C. [back_rewrite(20),rewrite([33(4)])].
% 0.75/1.02 44 inverse(divide(divide(A,B),inverse(B))) = inverse(A). [para(33(a,1),10(a,1,1,1,1))].
% 0.75/1.02 49 divide(A,divide(B,inverse(A))) = inverse(B). [para(33(a,1),32(a,1,2,1))].
% 0.75/1.02 51 divide(inverse(A),divide(B,A)) = inverse(B). [para(33(a,1),49(a,1,2,2))].
% 0.75/1.02 54 inverse(divide(A,inverse(divide(inverse(B),divide(C,divide(B,A)))))) = C. [para(49(a,1),34(a,1,1,2,1,1)),rewrite([33(3)])].
% 0.75/1.02 57 divide(inverse(divide(A,B)),inverse(A)) = B. [para(51(a,1),51(a,1,2)),rewrite([33(6)])].
% 0.75/1.02 58 divide(inverse(divide(inverse(A),B)),A) = B. [para(33(a,1),57(a,1,2))].
% 0.75/1.02 59 divide(divide(A,B),inverse(B)) = A. [para(44(a,1),33(a,1,1)),rewrite([33(2)]),flip(a)].
% 0.75/1.02 63 inverse(divide(A,B)) = divide(B,A). [para(57(a,1),44(a,1,1,1)),rewrite([33(2),33(5)])].
% 0.75/1.02 64 divide(divide(A,inverse(B)),B) = A. [para(44(a,1),58(a,1,1,1,1)),rewrite([63(3),59(5)])].
% 0.75/1.02 66 divide(divide(divide(A,divide(B,C)),inverse(B)),C) = A. [back_rewrite(54),rewrite([63(5),63(6)])].
% 0.75/1.02 70 divide(divide(a3,inverse(b3)),inverse(c3)) != divide(a3,divide(inverse(c3),b3)) # answer(prove_these_axioms_3). [back_rewrite(8),rewrite([63(13)])].
% 0.75/1.02 102 divide(divide(A,inverse(B)),C) = divide(A,divide(C,B)). [para(64(a,1),66(a,1,1,1)),rewrite([63(5)])].
% 0.75/1.02 103 $F # answer(prove_these_axioms_3). [resolve(102,a,70,a)].
% 0.75/1.02
% 0.75/1.02 % SZS output end Refutation
% 0.75/1.02 ============================== end of proof ==========================
% 0.75/1.02
% 0.75/1.02 ============================== STATISTICS ============================
% 0.75/1.02
% 0.75/1.02 Given=27. Generated=399. Kept=98. proofs=1.
% 0.75/1.02 Usable=14. Sos=25. Demods=46. Limbo=8, Disabled=54. Hints=0.
% 0.75/1.02 Megabytes=0.11.
% 0.75/1.02 User_CPU=0.02, System_CPU=0.00, Wall_clock=0.
% 0.75/1.02
% 0.75/1.02 ============================== end of statistics =====================
% 0.75/1.02
% 0.75/1.02 ============================== end of search =========================
% 0.75/1.02
% 0.75/1.02 THEOREM PROVED
% 0.75/1.02 % SZS status Unsatisfiable
% 0.75/1.02
% 0.75/1.02 Exiting with 1 proof.
% 0.75/1.02
% 0.75/1.02 Process 25635 exit (max_proofs) Tue Jun 14 02:32:05 2022
% 0.75/1.02 Prover9 interrupted
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