TSTP Solution File: SWV415+1 by SOS---2.0
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- Process Solution
%------------------------------------------------------------------------------
% File : SOS---2.0
% Problem : SWV415+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm : none
% Format : tptp:raw
% Command : sos-script %s
% Computer : n017.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 : Wed Jul 20 21:38:23 EDT 2022
% Result : Theorem 0.40s 0.62s
% Output : Refutation 0.40s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.13/0.13 % Problem : SWV415+1 : TPTP v8.1.0. Released v3.3.0.
% 0.13/0.14 % Command : sos-script %s
% 0.14/0.36 % Computer : n017.cluster.edu
% 0.14/0.36 % Model : x86_64 x86_64
% 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36 % Memory : 8042.1875MB
% 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36 % CPULimit : 300
% 0.14/0.36 % WCLimit : 600
% 0.14/0.36 % DateTime : Wed Jun 15 21:25:10 EDT 2022
% 0.14/0.36 % CPUTime :
% 0.14/0.38 ----- Otter 3.2, August 2001 -----
% 0.14/0.38 The process was started by sandbox2 on n017.cluster.edu,
% 0.14/0.38 Wed Jun 15 21:25:10 2022
% 0.14/0.38 The command was "./sos". The process ID is 15991.
% 0.14/0.38
% 0.14/0.38 set(prolog_style_variables).
% 0.14/0.38 set(auto).
% 0.14/0.38 dependent: set(auto1).
% 0.14/0.38 dependent: set(process_input).
% 0.14/0.38 dependent: clear(print_kept).
% 0.14/0.38 dependent: clear(print_new_demod).
% 0.14/0.38 dependent: clear(print_back_demod).
% 0.14/0.38 dependent: clear(print_back_sub).
% 0.14/0.38 dependent: set(control_memory).
% 0.14/0.38 dependent: assign(max_mem, 12000).
% 0.14/0.38 dependent: assign(pick_given_ratio, 4).
% 0.14/0.38 dependent: assign(stats_level, 1).
% 0.14/0.38 dependent: assign(pick_semantic_ratio, 3).
% 0.14/0.38 dependent: assign(sos_limit, 5000).
% 0.14/0.38 dependent: assign(max_weight, 60).
% 0.14/0.38 clear(print_given).
% 0.14/0.38
% 0.14/0.38 formula_list(usable).
% 0.14/0.38
% 0.14/0.38 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=4.
% 0.14/0.38
% 0.14/0.38 This ia a non-Horn set with equality. The strategy will be
% 0.14/0.38 Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.14/0.38 unit deletion, with positive clauses in sos and nonpositive
% 0.14/0.38 clauses in usable.
% 0.14/0.38
% 0.14/0.38 dependent: set(knuth_bendix).
% 0.14/0.38 dependent: set(para_from).
% 0.14/0.38 dependent: set(para_into).
% 0.14/0.38 dependent: clear(para_from_right).
% 0.14/0.38 dependent: clear(para_into_right).
% 0.14/0.38 dependent: set(para_from_vars).
% 0.14/0.38 dependent: set(eq_units_both_ways).
% 0.14/0.38 dependent: set(dynamic_demod_all).
% 0.14/0.38 dependent: set(dynamic_demod).
% 0.14/0.38 dependent: set(order_eq).
% 0.14/0.38 dependent: set(back_demod).
% 0.14/0.38 dependent: set(lrpo).
% 0.14/0.38 dependent: set(hyper_res).
% 0.14/0.38 dependent: set(unit_deletion).
% 0.14/0.38 dependent: set(factor).
% 0.14/0.38
% 0.14/0.38 ------------> process usable:
% 0.14/0.38
% 0.14/0.38 ------------> process sos:
% 0.14/0.38 Following clause subsumed by 74 during input processing: 0 [copy,74,flip.1] {-} insert_pq(insert_pq(A,B),C)=insert_pq(insert_pq(A,C),B).
% 0.14/0.38 Following clause subsumed by 101 during input processing: 0 [copy,101,flip.1] {-} i(triple(A,B,C))=i(triple(D,B,E)).
% 0.14/0.38 Following clause subsumed by 102 during input processing: 0 [copy,102,flip.1] {-} A=A.
% 0.14/0.38 Following clause subsumed by 85 during input processing: 0 [copy,103,flip.1] {-} insert_cpq(triple(A,B,C),D)=triple(insert_pqp(A,D),insert_slb(B,pair(D,bottom)),C).
% 0.14/0.38
% 0.14/0.38 ======= end of input processing =======
% 0.21/0.46
% 0.21/0.46
% 0.21/0.46 Failed to model usable list: disabling FINDER
% 0.21/0.46
% 0.21/0.46
% 0.21/0.46
% 0.21/0.46 -------------- Softie stats --------------
% 0.21/0.46
% 0.21/0.46 UPDATE_STOP: 300
% 0.21/0.46 SFINDER_TIME_LIMIT: 2
% 0.21/0.46 SHORT_CLAUSE_CUTOFF: 4
% 0.21/0.46 number of clauses in intial UL: 63
% 0.21/0.46 number of clauses initially in problem: 101
% 0.21/0.46 percentage of clauses intially in UL: 62
% 0.21/0.46 percentage of distinct symbols occuring in initial UL: 90
% 0.21/0.46 percent of all initial clauses that are short: 100
% 0.21/0.46 absolute distinct symbol count: 54
% 0.21/0.46 distinct predicate count: 20
% 0.21/0.46 distinct function count: 26
% 0.21/0.46 distinct constant count: 8
% 0.21/0.46
% 0.21/0.46 ---------- no more Softie stats ----------
% 0.21/0.46
% 0.21/0.46
% 0.21/0.46
% 0.21/0.46 =========== start of search ===========
% 0.40/0.62
% 0.40/0.62 -------- PROOF --------
% 0.40/0.62 % SZS status Theorem
% 0.40/0.62 % SZS output start Refutation
% 0.40/0.62
% 0.40/0.62 ----> UNIT CONFLICT at 0.21 sec ----> 2428 [binary,2427.1,995.1] {-} $F.
% 0.40/0.62
% 0.40/0.62 Length of proof is 2. Level of proof is 1.
% 0.40/0.62
% 0.40/0.62 ---------------- PROOF ----------------
% 0.40/0.62 % SZS status Theorem
% 0.40/0.62 % SZS output start Refutation
% 0.40/0.62
% 0.40/0.62 65 [] {+} i(insert_cpq(triple($c4,$c3,$c2),$c1))!=insert_pq(i(triple($c4,$c3,$c2)),$c1).
% 0.40/0.62 85 [] {-} insert_cpq(triple(A,B,C),D)=triple(insert_pqp(A,D),insert_slb(B,pair(D,bottom)),C).
% 0.40/0.62 100,99 [] {-} i(triple(A,insert_slb(B,pair(C,D)),E))=insert_pq(i(triple(A,B,E)),C).
% 0.40/0.62 101 [] {-} i(triple(A,B,C))=i(triple(D,B,E)).
% 0.40/0.62 995 [para_from,85.1.1,65.1.1.1,demod,100] {-} insert_pq(i(triple(insert_pqp($c4,$c1),$c3,$c2)),$c1)!=insert_pq(i(triple($c4,$c3,$c2)),$c1).
% 0.40/0.62 2427 [para_into,101.1.1,99.1.1,demod,100] {-} insert_pq(i(triple(A,B,C)),D)=insert_pq(i(triple(E,B,F)),D).
% 0.40/0.62 2428 [binary,2427.1,995.1] {-} $F.
% 0.40/0.62
% 0.40/0.62 % SZS output end Refutation
% 0.40/0.62 ------------ end of proof -------------
% 0.40/0.62
% 0.40/0.62
% 0.40/0.62 Search stopped by max_proofs option.
% 0.40/0.62
% 0.40/0.62
% 0.40/0.62 Search stopped by max_proofs option.
% 0.40/0.62
% 0.40/0.62 ============ end of search ============
% 0.40/0.62
% 0.40/0.62 That finishes the proof of the theorem.
% 0.40/0.62
% 0.40/0.62 Process 15991 finished Wed Jun 15 21:25:10 2022
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