TSTP Solution File: HEN005-5 by Prover9---1109a
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- Process Solution
%------------------------------------------------------------------------------
% File : Prover9---1109a
% Problem : HEN005-5 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : tptp2X_and_run_prover9 %d %s
% Computer : n021.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 13:00:48 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
% 0.07/0.12 % Problem : HEN005-5 : TPTP v8.1.0. Released v1.0.0.
% 0.07/0.13 % Command : tptp2X_and_run_prover9 %d %s
% 0.12/0.35 % Computer : n021.cluster.edu
% 0.12/0.35 % Model : x86_64 x86_64
% 0.12/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.35 % Memory : 8042.1875MB
% 0.12/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.35 % CPULimit : 300
% 0.12/0.35 % WCLimit : 600
% 0.12/0.35 % DateTime : Fri Jul 1 13:10:59 EDT 2022
% 0.12/0.35 % CPUTime :
% 0.75/1.02 ============================== Prover9 ===============================
% 0.75/1.02 Prover9 (32) version 2009-11A, November 2009.
% 0.75/1.02 Process 24651 was started by sandbox on n021.cluster.edu,
% 0.75/1.02 Fri Jul 1 13:10:59 2022
% 0.75/1.02 The command was "/export/starexec/sandbox/solver/bin/prover9 -t 300 -f /tmp/Prover9_24498_n021.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_24498_n021.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 (8 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_transitivity_of_divide_to_zero to answer in negative clause
% 0.75/1.02
% 0.75/1.02 Term ordering decisions:
% 0.75/1.02 Function symbol KB weights: zero=1. b=1. a=1. c=1. identity=1. divide=1.
% 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_transitivity_of_divide_to_zero.
% 0.75/1.02 % Length of proof is 31.
% 0.75/1.02 % Level of proof is 10.
% 0.75/1.02 % Maximum clause weight is 17.000.
% 0.75/1.02 % Given clauses 18.
% 0.75/1.02
% 0.75/1.02 1 divide(zero,A) = zero # label(zero_is_smallest) # label(axiom). [assumption].
% 0.75/1.02 2 divide(A,identity) = zero # label(identity_is_largest) # label(axiom). [assumption].
% 0.75/1.02 3 divide(a,b) = zero # label(a_divide_b_is_zero) # label(hypothesis). [assumption].
% 0.75/1.02 4 zero = divide(a,b). [copy(3),flip(a)].
% 0.75/1.02 5 divide(b,c) = zero # label(b_divide_c_is_zero) # label(hypothesis). [assumption].
% 0.75/1.02 6 divide(a,b) = divide(b,c). [copy(5),rewrite([4(4)]),flip(a)].
% 0.75/1.02 9 divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)) = zero # label(quotient_property) # label(axiom). [assumption].
% 0.75/1.02 10 divide(divide(divide(A,B),divide(C,B)),divide(divide(A,C),B)) = divide(b,c). [copy(9),rewrite([4(7),6(9)])].
% 0.75/1.02 11 divide(a,c) != zero # label(prove_transitivity_of_divide_to_zero) # label(negated_conjecture) # answer(prove_transitivity_of_divide_to_zero). [assumption].
% 0.75/1.02 12 divide(a,c) != divide(b,c) # answer(prove_transitivity_of_divide_to_zero). [copy(11),rewrite([4(4),6(6)])].
% 0.75/1.02 13 divide(A,B) != zero | divide(B,A) != zero | A = B # label(divide_and_equal) # label(axiom). [assumption].
% 0.75/1.02 14 divide(b,c) != divide(A,B) | divide(b,c) != divide(B,A) | A = B. [copy(13),rewrite([4(2),6(4),4(7),6(9)]),flip(a),flip(b)].
% 0.75/1.02 15 divide(b,c) = divide(A,identity). [back_rewrite(2),rewrite([4(3),6(5)]),flip(a)].
% 0.75/1.02 16 divide(divide(b,c),A) = divide(b,c). [back_rewrite(1),rewrite([4(1),6(3),4(5),6(7)])].
% 0.75/1.02 18 divide(a,c) = c_0. [new_symbol(12)].
% 0.75/1.02 19 divide(b,c) != c_0 # answer(prove_transitivity_of_divide_to_zero). [back_rewrite(12),rewrite([18(3)]),flip(a)].
% 0.75/1.02 33 divide(divide(divide(A,c),divide(B,identity)),divide(divide(A,b),c)) = divide(b,c). [para(15(a,1),10(a,1,1,2))].
% 0.75/1.02 38 divide(A,identity) != divide(B,C) | divide(b,c) != divide(C,B) | B = C. [para(15(a,1),14(a,1))].
% 0.75/1.02 42 divide(A,identity) = divide(B,identity). [para(15(a,1),15(a,1))].
% 0.75/1.02 43 divide(A,identity) = c_1. [new_symbol(42)].
% 0.75/1.02 46 divide(A,B) != c_1 | divide(b,c) != divide(B,A) | A = B. [back_rewrite(38),rewrite([43(2)]),flip(a)].
% 0.75/1.02 50 divide(divide(divide(A,c),c_1),divide(divide(A,b),c)) = divide(b,c). [back_rewrite(33),rewrite([43(4)])].
% 0.75/1.02 53 divide(b,c) = c_1. [back_rewrite(15),rewrite([43(5)])].
% 0.75/1.02 56 divide(divide(divide(A,c),c_1),divide(divide(A,b),c)) = c_1. [back_rewrite(50),rewrite([53(12)])].
% 0.75/1.02 58 divide(A,B) != c_1 | divide(B,A) != c_1 | A = B. [back_rewrite(46),rewrite([53(6)]),flip(b)].
% 0.75/1.02 73 c_1 != c_0 # answer(prove_transitivity_of_divide_to_zero). [back_rewrite(19),rewrite([53(3)])].
% 0.75/1.02 75 divide(c_1,A) = c_1. [back_rewrite(16),rewrite([53(3),53(5)])].
% 0.75/1.02 78 divide(a,b) = c_1. [back_rewrite(6),rewrite([53(6)])].
% 0.75/1.02 80 divide(divide(c_0,c_1),c_1) = c_1. [para(18(a,1),56(a,1,1,1)),rewrite([78(6),75(6)])].
% 0.75/1.02 84 divide(c_0,c_1) = c_1. [hyper(58,a,80,a,b,75,a)].
% 0.75/1.02 85 $F # answer(prove_transitivity_of_divide_to_zero). [ur(58,b,75,a,c,73,a(flip)),rewrite([84(3)]),xx(a)].
% 0.75/1.02
% 0.75/1.02 % SZS output end Refutation
% 0.75/1.02 ============================== end of proof ==========================
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% 0.75/1.02 ============================== STATISTICS ============================
% 0.75/1.02
% 0.75/1.02 Given=18. Generated=146. Kept=78. proofs=1.
% 0.75/1.02 Usable=13. Sos=17. Demods=24. Limbo=1, Disabled=55. Hints=0.
% 0.75/1.02 Megabytes=0.09.
% 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 24651 exit (max_proofs) Fri Jul 1 13:10:59 2022
% 0.75/1.02 Prover9 interrupted
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