TSTP Solution File: SET676+3 by SOS---2.0
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- Process Solution
%------------------------------------------------------------------------------
% File : SOS---2.0
% Problem : SET676+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm : none
% Format : tptp:raw
% Command : sos-script %s
% Computer : n005.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 : Tue Jul 19 05:20:00 EDT 2022
% Result : Theorem 0.99s 1.20s
% Output : Refutation 0.99s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SET676+3 : TPTP v8.1.0. Released v2.2.0.
% 0.07/0.13 % Command : sos-script %s
% 0.13/0.34 % Computer : n005.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 : Sun Jul 10 12:22:52 EDT 2022
% 0.13/0.34 % CPUTime :
% 0.20/0.39 ----- Otter 3.2, August 2001 -----
% 0.20/0.39 The process was started by sandbox on n005.cluster.edu,
% 0.20/0.39 Sun Jul 10 12:22:52 2022
% 0.20/0.39 The command was "./sos". The process ID is 24294.
% 0.20/0.39
% 0.20/0.39 set(prolog_style_variables).
% 0.20/0.39 set(auto).
% 0.20/0.39 dependent: set(auto1).
% 0.20/0.39 dependent: set(process_input).
% 0.20/0.39 dependent: clear(print_kept).
% 0.20/0.39 dependent: clear(print_new_demod).
% 0.20/0.39 dependent: clear(print_back_demod).
% 0.20/0.39 dependent: clear(print_back_sub).
% 0.20/0.39 dependent: set(control_memory).
% 0.20/0.39 dependent: assign(max_mem, 12000).
% 0.20/0.39 dependent: assign(pick_given_ratio, 4).
% 0.20/0.39 dependent: assign(stats_level, 1).
% 0.20/0.39 dependent: assign(pick_semantic_ratio, 3).
% 0.20/0.39 dependent: assign(sos_limit, 5000).
% 0.20/0.39 dependent: assign(max_weight, 60).
% 0.20/0.39 clear(print_given).
% 0.20/0.39
% 0.20/0.39 formula_list(usable).
% 0.20/0.39
% 0.20/0.39 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=9.
% 0.20/0.39
% 0.20/0.39 This ia a non-Horn set with equality. The strategy will be
% 0.20/0.39 Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.20/0.39 unit deletion, with positive clauses in sos and nonpositive
% 0.20/0.39 clauses in usable.
% 0.20/0.39
% 0.20/0.39 dependent: set(knuth_bendix).
% 0.20/0.39 dependent: set(para_from).
% 0.20/0.39 dependent: set(para_into).
% 0.20/0.39 dependent: clear(para_from_right).
% 0.20/0.39 dependent: clear(para_into_right).
% 0.20/0.39 dependent: set(para_from_vars).
% 0.20/0.39 dependent: set(eq_units_both_ways).
% 0.20/0.39 dependent: set(dynamic_demod_all).
% 0.20/0.39 dependent: set(dynamic_demod).
% 0.20/0.39 dependent: set(order_eq).
% 0.20/0.39 dependent: set(back_demod).
% 0.20/0.39 dependent: set(lrpo).
% 0.20/0.39 dependent: set(hyper_res).
% 0.20/0.39 dependent: set(unit_deletion).
% 0.20/0.39 dependent: set(factor).
% 0.20/0.39
% 0.20/0.39 ------------> process usable:
% 0.20/0.39
% 0.20/0.39 ------------> process sos:
% 0.20/0.39 Following clause subsumed by 140 during input processing: 0 [] {-} ilf_type($c1,set_type).
% 0.20/0.39 140 back subsumes 95.
% 0.20/0.39 140 back subsumes 94.
% 0.20/0.39 140 back subsumes 88.
% 0.20/0.39 140 back subsumes 87.
% 0.20/0.39 140 back subsumes 81.
% 0.20/0.39 140 back subsumes 72.
% 0.20/0.39 140 back subsumes 69.
% 0.20/0.39 140 back subsumes 48.
% 0.20/0.39 140 back subsumes 47.
% 0.20/0.39 140 back subsumes 46.
% 0.20/0.39 140 back subsumes 45.
% 0.20/0.39 140 back subsumes 44.
% 0.20/0.39 140 back subsumes 43.
% 0.20/0.39 140 back subsumes 36.
% 0.20/0.39 140 back subsumes 33.
% 0.20/0.39 140 back subsumes 32.
% 0.20/0.39 140 back subsumes 30.
% 0.20/0.39 140 back subsumes 25.
% 0.20/0.39 140 back subsumes 21.
% 0.20/0.39 140 back subsumes 13.
% 0.20/0.39 140 back subsumes 9.
% 0.20/0.39 140 back subsumes 3.
% 0.20/0.39 140 back subsumes 2.
% 0.20/0.39 Following clause subsumed by 141 during input processing: 0 [copy,141,flip.1] {-} A=A.
% 0.20/0.39
% 0.20/0.39 ======= end of input processing =======
% 0.38/0.60
% 0.38/0.60 Model 1 (0.00 seconds, 0 Inserts)
% 0.38/0.60
% 0.38/0.60 Stopped by limit on number of solutions
% 0.38/0.60
% 0.38/0.60
% 0.38/0.60 -------------- Softie stats --------------
% 0.38/0.60
% 0.38/0.60 UPDATE_STOP: 300
% 0.38/0.60 SFINDER_TIME_LIMIT: 2
% 0.38/0.60 SHORT_CLAUSE_CUTOFF: 4
% 0.38/0.60 number of clauses in intial UL: 114
% 0.38/0.60 number of clauses initially in problem: 116
% 0.38/0.60 percentage of clauses intially in UL: 98
% 0.38/0.60 percentage of distinct symbols occuring in initial UL: 100
% 0.38/0.60 percent of all initial clauses that are short: 100
% 0.38/0.60 absolute distinct symbol count: 25
% 0.38/0.60 distinct predicate count: 5
% 0.38/0.60 distinct function count: 18
% 0.38/0.60 distinct constant count: 2
% 0.38/0.60
% 0.38/0.60 ---------- no more Softie stats ----------
% 0.38/0.60
% 0.38/0.60
% 0.38/0.60
% 0.38/0.60 Stopped by limit on insertions
% 0.38/0.60
% 0.38/0.60 =========== start of search ===========
% 0.99/1.20
% 0.99/1.20 -------- PROOF --------
% 0.99/1.20 % SZS status Theorem
% 0.99/1.20 % SZS output start Refutation
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 2 [ 3 1 811 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 3 [ 2 0 3787 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 4 [ 1 2 13502 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 5 [ 3 0 933 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 6 [ 2 0 3597 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 7 [ 3 1 1581 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 Stopped by limit on insertions
% 0.99/1.20
% 0.99/1.20 Model 8 [ 5 0 2344 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 ----> UNIT CONFLICT at 0.82 sec ----> 174 [binary,173.1,41.1] {-} $F.
% 0.99/1.20
% 0.99/1.20 Length of proof is 2. Level of proof is 2.
% 0.99/1.20
% 0.99/1.20 ---------------- PROOF ----------------
% 0.99/1.20 % SZS status Theorem
% 0.99/1.20 % SZS output start Refutation
% 0.99/1.20
% 0.99/1.20 1 [] {+} -ilf_type(A,set_type)| -ilf_type(B,set_type)|ilf_type(cross_product(A,B),relation_type(A,B)).
% 0.99/1.20 11 [] {+} -ilf_type(A,set_type)| -ilf_type(B,set_type)|ilf_type(B,identity_relation_of_type(A))| -ilf_type(B,relation_type(A,A)).
% 0.99/1.20 41 [] {+} -ilf_type(cross_product($c1,$c1),identity_relation_of_type($c1)).
% 0.99/1.20 140 [] {-} ilf_type(A,set_type).
% 0.99/1.20 152 [hyper,140,1,140] {+} ilf_type(cross_product(A,B),relation_type(A,B)).
% 0.99/1.20 173 [hyper,152,11,140,140] {-} ilf_type(cross_product(A,A),identity_relation_of_type(A)).
% 0.99/1.20 174 [binary,173.1,41.1] {-} $F.
% 0.99/1.20
% 0.99/1.20 % SZS output end Refutation
% 0.99/1.20 ------------ end of proof -------------
% 0.99/1.20
% 0.99/1.20
% 0.99/1.20 Search stopped by max_proofs option.
% 0.99/1.20
% 0.99/1.20
% 0.99/1.20 Search stopped by max_proofs option.
% 0.99/1.20
% 0.99/1.20 ============ end of search ============
% 0.99/1.20
% 0.99/1.20 ----------- soft-scott stats ----------
% 0.99/1.20
% 0.99/1.20 true clauses given 3 (25.0%)
% 0.99/1.20 false clauses given 9
% 0.99/1.20
% 0.99/1.20 FALSE TRUE
% 0.99/1.20 6 0 3
% 0.99/1.20 7 2 0
% 0.99/1.20 8 1 2
% 0.99/1.20 9 1 0
% 0.99/1.20 10 0 1
% 0.99/1.20 13 1 0
% 0.99/1.20 23 0 1
% 0.99/1.20 tot: 5 7 (58.3% true)
% 0.99/1.20
% 0.99/1.20
% 0.99/1.20 Model 8 [ 5 0 2344 ] (0.00 seconds, 250000 Inserts)
% 0.99/1.20
% 0.99/1.20 That finishes the proof of the theorem.
% 0.99/1.20
% 0.99/1.20 Process 24294 finished Sun Jul 10 12:22:52 2022
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