TSTP Solution File: KRS070+1 by SOS---2.0
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : SOS---2.0
% Problem : KRS070+1 : TPTP v8.1.0. Released v3.1.0.
% Transfm : none
% Format : tptp:raw
% Command : sos-script %s
% Computer : n019.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 : Sun Jul 17 03:29:41 EDT 2022
% Result : Unsatisfiable 0.75s 1.02s
% Output : Refutation 0.75s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11 % Problem : KRS070+1 : TPTP v8.1.0. Released v3.1.0.
% 0.11/0.12 % Command : sos-script %s
% 0.12/0.33 % Computer : n019.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 : 600
% 0.12/0.33 % DateTime : Tue Jun 7 10:32:10 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.12/0.34 ----- Otter 3.2, August 2001 -----
% 0.12/0.34 The process was started by sandbox on n019.cluster.edu,
% 0.12/0.34 Tue Jun 7 10:32:10 2022
% 0.12/0.34 The command was "./sos". The process ID is 25481.
% 0.12/0.34
% 0.12/0.34 set(prolog_style_variables).
% 0.12/0.34 set(auto).
% 0.12/0.34 dependent: set(auto1).
% 0.12/0.34 dependent: set(process_input).
% 0.12/0.34 dependent: clear(print_kept).
% 0.12/0.34 dependent: clear(print_new_demod).
% 0.12/0.34 dependent: clear(print_back_demod).
% 0.12/0.34 dependent: clear(print_back_sub).
% 0.12/0.34 dependent: set(control_memory).
% 0.12/0.34 dependent: assign(max_mem, 12000).
% 0.12/0.34 dependent: assign(pick_given_ratio, 4).
% 0.12/0.34 dependent: assign(stats_level, 1).
% 0.12/0.34 dependent: assign(pick_semantic_ratio, 3).
% 0.12/0.34 dependent: assign(sos_limit, 5000).
% 0.12/0.34 dependent: assign(max_weight, 60).
% 0.12/0.34 clear(print_given).
% 0.12/0.34
% 0.12/0.34 formula_list(usable).
% 0.12/0.34
% 0.12/0.34 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=6.
% 0.12/0.34
% 0.12/0.34 This ia a non-Horn set with equality. The strategy will be
% 0.12/0.34 Knuth-Bendix, ordered hyper_res, ur_res, factoring, and
% 0.12/0.34 unit deletion, with positive clauses in sos and nonpositive
% 0.12/0.34 clauses in usable.
% 0.12/0.34
% 0.12/0.34 dependent: set(knuth_bendix).
% 0.12/0.34 dependent: set(para_from).
% 0.12/0.34 dependent: set(para_into).
% 0.12/0.34 dependent: clear(para_from_right).
% 0.12/0.34 dependent: clear(para_into_right).
% 0.12/0.34 dependent: set(para_from_vars).
% 0.12/0.34 dependent: set(eq_units_both_ways).
% 0.12/0.34 dependent: set(dynamic_demod_all).
% 0.12/0.34 dependent: set(dynamic_demod).
% 0.12/0.34 dependent: set(order_eq).
% 0.12/0.34 dependent: set(back_demod).
% 0.12/0.34 dependent: set(lrpo).
% 0.12/0.34 dependent: set(hyper_res).
% 0.12/0.34 dependent: set(unit_deletion).
% 0.12/0.34 dependent: set(factor).
% 0.12/0.34
% 0.12/0.34 ------------> process usable:
% 0.12/0.34 28 back subsumes 4.
% 0.12/0.34
% 0.12/0.34 ------------> process sos:
% 0.12/0.34 56 back subsumes 5.
% 0.12/0.34 Following clause subsumed by 59 during input processing: 0 [copy,59,flip.1] {-} A=A.
% 0.12/0.34 59 back subsumes 55.
% 0.12/0.34 59 back subsumes 54.
% 0.12/0.34 59 back subsumes 53.
% 0.12/0.34 59 back subsumes 52.
% 0.12/0.34 59 back subsumes 51.
% 0.12/0.34
% 0.12/0.34 ======= end of input processing =======
% 0.12/0.38
% 0.12/0.38 Model 1 (0.00 seconds, 0 Inserts)
% 0.12/0.38
% 0.12/0.38 Stopped by limit on number of solutions
% 0.12/0.38
% 0.12/0.38
% 0.12/0.38 -------------- Softie stats --------------
% 0.12/0.38
% 0.12/0.38 UPDATE_STOP: 300
% 0.12/0.38 SFINDER_TIME_LIMIT: 2
% 0.12/0.38 SHORT_CLAUSE_CUTOFF: 4
% 0.12/0.38 number of clauses in intial UL: 48
% 0.12/0.38 number of clauses initially in problem: 52
% 0.12/0.38 percentage of clauses intially in UL: 92
% 0.12/0.38 percentage of distinct symbols occuring in initial UL: 90
% 0.12/0.38 percent of all initial clauses that are short: 100
% 0.12/0.38 absolute distinct symbol count: 22
% 0.12/0.38 distinct predicate count: 18
% 0.12/0.38 distinct function count: 3
% 0.12/0.38 distinct constant count: 1
% 0.12/0.38
% 0.12/0.38 ---------- no more Softie stats ----------
% 0.12/0.38
% 0.12/0.38
% 0.12/0.38
% 0.12/0.38 Model 2 (0.00 seconds, 0 Inserts)
% 0.12/0.38
% 0.12/0.38 Stopped by limit on number of solutions
% 0.12/0.38
% 0.12/0.38 =========== start of search ===========
% 0.75/1.02
% 0.75/1.02 -------- PROOF --------
% 0.75/1.02 % SZS status Theorem
% 0.75/1.02 % SZS output start Refutation
% 0.75/1.02
% 0.75/1.02 Model 3 [ 1 1 254 ] (0.00 seconds, 5387 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 4 [ 1 1 282 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Model 5 [ 1 3 118093 ] (0.00 seconds, 124670 Inserts)
% 0.75/1.02
% 0.75/1.02 Model 6 [ 1 1 494 ] (0.00 seconds, 5117 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 7 [ 3 1 196 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Model 8 [ 1 1 544 ] (0.00 seconds, 6908 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 9 [ 3 1 216 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 10 [ 3 1 234 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 11 [ 2 1 255 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 12 [ 2 0 612 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 Stopped by limit on insertions
% 0.75/1.02
% 0.75/1.02 Model 13 [ 2 6 135888 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 -----> EMPTY CLAUSE at 0.65 sec ----> 223 [hyper,207,32,58,206,62] {-} $F.
% 0.75/1.02
% 0.75/1.02 Length of proof is 9. Level of proof is 4.
% 0.75/1.02
% 0.75/1.02 ---------------- PROOF ----------------
% 0.75/1.02 % SZS status Theorem
% 0.75/1.02 % SZS output start Refutation
% 0.75/1.02
% 0.75/1.02 30 [] {+} -cUnsatisfiable(A)|rrx4(A,$f1(A)).
% 0.75/1.02 31 [] {+} -cUnsatisfiable(A)|cc2($f1(A)).
% 0.75/1.02 32 [] {+} -cUnsatisfiable(A)| -rrx3(A,B)| -cc2(B)| -cc1(B).
% 0.75/1.02 33 [] {+} -cUnsatisfiable(A)|rrx3(A,$f2(A)).
% 0.75/1.02 34 [] {+} -cUnsatisfiable(A)|cc1($f2(A)).
% 0.75/1.02 38 [] {+} -rrx(A,B)| -rrx(A,C)|B=C.
% 0.75/1.02 43 [] {+} -rrx3(A,B)|rrx(A,B).
% 0.75/1.02 50 [] {+} -rrx4(A,B)|rrx(A,B).
% 0.75/1.02 58 [] {-} cUnsatisfiable(i2003_11_14_17_18_32242).
% 0.75/1.02 60 [hyper,58,34] {+} cc1($f2(i2003_11_14_17_18_32242)).
% 0.75/1.02 61 [hyper,58,33] {+} rrx3(i2003_11_14_17_18_32242,$f2(i2003_11_14_17_18_32242)).
% 0.75/1.02 62 [hyper,58,31] {+} cc2($f1(i2003_11_14_17_18_32242)).
% 0.75/1.02 63 [hyper,58,30] {+} rrx4(i2003_11_14_17_18_32242,$f1(i2003_11_14_17_18_32242)).
% 0.75/1.02 80 [hyper,61,43] {+} rrx(i2003_11_14_17_18_32242,$f2(i2003_11_14_17_18_32242)).
% 0.75/1.02 96 [hyper,63,50] {+} rrx(i2003_11_14_17_18_32242,$f1(i2003_11_14_17_18_32242)).
% 0.75/1.02 169,168 [hyper,96,38,80] {+} $f2(i2003_11_14_17_18_32242)=$f1(i2003_11_14_17_18_32242).
% 0.75/1.02 206 [back_demod,61,demod,169] {-} rrx3(i2003_11_14_17_18_32242,$f1(i2003_11_14_17_18_32242)).
% 0.75/1.02 207 [back_demod,60,demod,169] {+} cc1($f1(i2003_11_14_17_18_32242)).
% 0.75/1.02 223 [hyper,207,32,58,206,62] {-} $F.
% 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
% 0.75/1.02 Search stopped by max_proofs option.
% 0.75/1.02
% 0.75/1.02
% 0.75/1.02 Search stopped by max_proofs option.
% 0.75/1.02
% 0.75/1.02 ============ end of search ============
% 0.75/1.02
% 0.75/1.02 ----------- soft-scott stats ----------
% 0.75/1.02
% 0.75/1.02 true clauses given 4 (23.5%)
% 0.75/1.02 false clauses given 13
% 0.75/1.02
% 0.75/1.02 FALSE TRUE
% 0.75/1.02 4 0 4
% 0.75/1.02 5 0 1
% 0.75/1.02 6 0 5
% 0.75/1.02 7 0 2
% 0.75/1.02 8 1 0
% 0.75/1.02 9 0 25
% 0.75/1.02 10 6 94
% 0.75/1.02 14 0 20
% 0.75/1.02 tot: 7 151 (95.6% true)
% 0.75/1.02
% 0.75/1.02
% 0.75/1.02 Model 13 [ 2 6 135888 ] (0.00 seconds, 250000 Inserts)
% 0.75/1.02
% 0.75/1.02 That finishes the proof of the theorem.
% 0.75/1.02
% 0.75/1.02 Process 25481 finished Tue Jun 7 10:32:10 2022
%------------------------------------------------------------------------------