TSTP Solution File: SET596+3 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET596+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n012.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 : 300s
% DateTime : Wed Jul 27 13:13:47 EDT 2022
% Result : Theorem 1.70s 1.90s
% Output : Refutation 1.70s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 7
% Syntax : Number of clauses : 13 ( 11 unt; 0 nHn; 9 RR)
% Number of literals : 15 ( 6 equ; 6 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 4 con; 0-2 aty)
% Number of variables : 8 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ subset(A,empty_set)
| A = empty_set ),
file('SET596+3.p',unknown),
[] ).
cnf(2,axiom,
( ~ subset(A,B)
| subset(intersection(A,C),intersection(B,C)) ),
file('SET596+3.p',unknown),
[] ).
cnf(16,axiom,
intersection(dollar_c3,dollar_c1) != empty_set,
file('SET596+3.p',unknown),
[] ).
cnf(20,axiom,
A = A,
file('SET596+3.p',unknown),
[] ).
cnf(22,axiom,
intersection(A,B) = intersection(B,A),
file('SET596+3.p',unknown),
[] ).
cnf(26,axiom,
subset(dollar_c3,dollar_c2),
file('SET596+3.p',unknown),
[] ).
cnf(27,axiom,
intersection(dollar_c2,dollar_c1) = empty_set,
file('SET596+3.p',unknown),
[] ).
cnf(29,plain,
subset(intersection(dollar_c3,A),intersection(dollar_c2,A)),
inference(hyper,[status(thm)],[26,2]),
[iquote('hyper,26,2')] ).
cnf(81,plain,
intersection(dollar_c1,dollar_c3) != empty_set,
inference(para_from,[status(thm),theory(equality)],[22,16]),
[iquote('para_from,22.1.1,16.1.1')] ).
cnf(174,plain,
~ subset(intersection(dollar_c1,dollar_c3),empty_set),
inference(unit_del,[status(thm)],[inference(para_into,[status(thm),theory(equality)],[81,1]),20]),
[iquote('para_into,81.1.1,1.2.1,unit_del,20')] ).
cnf(185,plain,
~ subset(intersection(dollar_c3,dollar_c1),empty_set),
inference(para_into,[status(thm),theory(equality)],[174,22]),
[iquote('para_into,174.1.1,22.1.1')] ).
cnf(223,plain,
subset(intersection(dollar_c3,dollar_c1),empty_set),
inference(para_into,[status(thm),theory(equality)],[29,27]),
[iquote('para_into,29.1.2,27.1.1')] ).
cnf(224,plain,
$false,
inference(binary,[status(thm)],[223,185]),
[iquote('binary,223.1,185.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.11 % Problem : SET596+3 : TPTP v8.1.0. Released v2.2.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n012.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 : 300
% 0.12/0.33 % DateTime : Wed Jul 27 10:44:05 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.70/1.89 ----- Otter 3.3f, August 2004 -----
% 1.70/1.89 The process was started by sandbox2 on n012.cluster.edu,
% 1.70/1.89 Wed Jul 27 10:44:05 2022
% 1.70/1.89 The command was "./otter". The process ID is 25734.
% 1.70/1.89
% 1.70/1.89 set(prolog_style_variables).
% 1.70/1.89 set(auto).
% 1.70/1.89 dependent: set(auto1).
% 1.70/1.89 dependent: set(process_input).
% 1.70/1.89 dependent: clear(print_kept).
% 1.70/1.89 dependent: clear(print_new_demod).
% 1.70/1.89 dependent: clear(print_back_demod).
% 1.70/1.89 dependent: clear(print_back_sub).
% 1.70/1.89 dependent: set(control_memory).
% 1.70/1.89 dependent: assign(max_mem, 12000).
% 1.70/1.89 dependent: assign(pick_given_ratio, 4).
% 1.70/1.89 dependent: assign(stats_level, 1).
% 1.70/1.89 dependent: assign(max_seconds, 10800).
% 1.70/1.89 clear(print_given).
% 1.70/1.89
% 1.70/1.89 formula_list(usable).
% 1.70/1.89 all A (A=A).
% 1.70/1.89 all B (subset(B,empty_set)->B=empty_set).
% 1.70/1.89 all B C D (subset(B,C)->subset(intersection(B,D),intersection(C,D))).
% 1.70/1.89 all B (-member(B,empty_set)).
% 1.70/1.89 all B C D (member(D,intersection(B,C))<->member(D,B)&member(D,C)).
% 1.70/1.89 all B C (subset(B,C)<-> (all D (member(D,B)->member(D,C)))).
% 1.70/1.89 all B C (B=C<->subset(B,C)&subset(C,B)).
% 1.70/1.89 all B C (intersection(B,C)=intersection(C,B)).
% 1.70/1.89 all B subset(B,B).
% 1.70/1.89 all B (empty(B)<-> (all C (-member(C,B)))).
% 1.70/1.89 all B C (B=C<-> (all D (member(D,B)<->member(D,C)))).
% 1.70/1.89 -(all B C D (subset(B,C)&intersection(C,D)=empty_set->intersection(B,D)=empty_set)).
% 1.70/1.89 end_of_list.
% 1.70/1.89
% 1.70/1.89 -------> usable clausifies to:
% 1.70/1.89
% 1.70/1.89 list(usable).
% 1.70/1.89 0 [] A=A.
% 1.70/1.89 0 [] -subset(B,empty_set)|B=empty_set.
% 1.70/1.89 0 [] -subset(B,C)|subset(intersection(B,D),intersection(C,D)).
% 1.70/1.89 0 [] -member(B,empty_set).
% 1.70/1.89 0 [] -member(D,intersection(B,C))|member(D,B).
% 1.70/1.89 0 [] -member(D,intersection(B,C))|member(D,C).
% 1.70/1.89 0 [] member(D,intersection(B,C))| -member(D,B)| -member(D,C).
% 1.70/1.89 0 [] -subset(B,C)| -member(D,B)|member(D,C).
% 1.70/1.89 0 [] subset(B,C)|member($f1(B,C),B).
% 1.70/1.89 0 [] subset(B,C)| -member($f1(B,C),C).
% 1.70/1.89 0 [] B!=C|subset(B,C).
% 1.70/1.89 0 [] B!=C|subset(C,B).
% 1.70/1.89 0 [] B=C| -subset(B,C)| -subset(C,B).
% 1.70/1.89 0 [] intersection(B,C)=intersection(C,B).
% 1.70/1.89 0 [] subset(B,B).
% 1.70/1.89 0 [] -empty(B)| -member(C,B).
% 1.70/1.89 0 [] empty(B)|member($f2(B),B).
% 1.70/1.89 0 [] B!=C| -member(D,B)|member(D,C).
% 1.70/1.89 0 [] B!=C|member(D,B)| -member(D,C).
% 1.70/1.89 0 [] B=C|member($f3(B,C),B)|member($f3(B,C),C).
% 1.70/1.89 0 [] B=C| -member($f3(B,C),B)| -member($f3(B,C),C).
% 1.70/1.89 0 [] subset($c3,$c2).
% 1.70/1.89 0 [] intersection($c2,$c1)=empty_set.
% 1.70/1.89 0 [] intersection($c3,$c1)!=empty_set.
% 1.70/1.89 end_of_list.
% 1.70/1.89
% 1.70/1.89 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.70/1.89
% 1.70/1.89 This ia a non-Horn set with equality. The strategy will be
% 1.70/1.89 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.70/1.89 deletion, with positive clauses in sos and nonpositive
% 1.70/1.89 clauses in usable.
% 1.70/1.89
% 1.70/1.89 dependent: set(knuth_bendix).
% 1.70/1.89 dependent: set(anl_eq).
% 1.70/1.89 dependent: set(para_from).
% 1.70/1.89 dependent: set(para_into).
% 1.70/1.89 dependent: clear(para_from_right).
% 1.70/1.89 dependent: clear(para_into_right).
% 1.70/1.89 dependent: set(para_from_vars).
% 1.70/1.89 dependent: set(eq_units_both_ways).
% 1.70/1.89 dependent: set(dynamic_demod_all).
% 1.70/1.89 dependent: set(dynamic_demod).
% 1.70/1.89 dependent: set(order_eq).
% 1.70/1.89 dependent: set(back_demod).
% 1.70/1.89 dependent: set(lrpo).
% 1.70/1.89 dependent: set(hyper_res).
% 1.70/1.89 dependent: set(unit_deletion).
% 1.70/1.89 dependent: set(factor).
% 1.70/1.89
% 1.70/1.89 ------------> process usable:
% 1.70/1.89 ** KEPT (pick-wt=6): 1 [] -subset(A,empty_set)|A=empty_set.
% 1.70/1.89 ** KEPT (pick-wt=10): 2 [] -subset(A,B)|subset(intersection(A,C),intersection(B,C)).
% 1.70/1.89 ** KEPT (pick-wt=3): 3 [] -member(A,empty_set).
% 1.70/1.89 ** KEPT (pick-wt=8): 4 [] -member(A,intersection(B,C))|member(A,B).
% 1.70/1.89 ** KEPT (pick-wt=8): 5 [] -member(A,intersection(B,C))|member(A,C).
% 1.70/1.89 ** KEPT (pick-wt=11): 6 [] member(A,intersection(B,C))| -member(A,B)| -member(A,C).
% 1.70/1.89 ** KEPT (pick-wt=9): 7 [] -subset(A,B)| -member(C,A)|member(C,B).
% 1.70/1.89 ** KEPT (pick-wt=8): 8 [] subset(A,B)| -member($f1(A,B),B).
% 1.70/1.89 ** KEPT (pick-wt=6): 9 [] A!=B|subset(A,B).
% 1.70/1.89 ** KEPT (pick-wt=6): 10 [] A!=B|subset(B,A).
% 1.70/1.89 ** KEPT (pick-wt=9): 11 [] A=B| -subset(A,B)| -subset(B,A).
% 1.70/1.89 ** KEPT (pick-wt=5): 12 [] -empty(A)| -member(B,A).
% 1.70/1.89 ** KEPT (pick-wt=9): 13 [] A!=B| -member(C,A)|member(C,B).
% 1.70/1.89 ** KEPT (pick-wt=9): 14 [] A!=B|member(C,A)| -member(C,B).
% 1.70/1.89 ** KEPT (pick-wt=13): 15 [] A=B| -member($f3(A,B),A)| -member($f3(A,B),B).
% 1.70/1.89 ** KEPT (pick-wt=5): 16 [] intersection($c3,$c1)!=empty_set.
% 1.70/1.89
% 1.70/1.89 ------------> process sos:
% 1.70/1.89 ** KEPT (pick-wt=3): 20 [] A=A.
% 1.70/1.89 ** KEPT (pick-wt=8): 21 [] subset(A,B)|member($f1(A,B),A).
% 1.70/1.90 ** KEPT (pick-wt=7): 22 [] intersection(A,B)=intersection(B,A).
% 1.70/1.90 ** KEPT (pick-wt=3): 23 [] subset(A,A).
% 1.70/1.90 ** KEPT (pick-wt=6): 24 [] empty(A)|member($f2(A),A).
% 1.70/1.90 ** KEPT (pick-wt=13): 25 [] A=B|member($f3(A,B),A)|member($f3(A,B),B).
% 1.70/1.90 ** KEPT (pick-wt=3): 26 [] subset($c3,$c2).
% 1.70/1.90 ** KEPT (pick-wt=5): 27 [] intersection($c2,$c1)=empty_set.
% 1.70/1.90 ---> New Demodulator: 28 [new_demod,27] intersection($c2,$c1)=empty_set.
% 1.70/1.90 Following clause subsumed by 20 during input processing: 0 [copy,20,flip.1] A=A.
% 1.70/1.90 20 back subsumes 19.
% 1.70/1.90 20 back subsumes 18.
% 1.70/1.90 Following clause subsumed by 22 during input processing: 0 [copy,22,flip.1] intersection(A,B)=intersection(B,A).
% 1.70/1.90 >>>> Starting back demodulation with 28.
% 1.70/1.90
% 1.70/1.90 ======= end of input processing =======
% 1.70/1.90
% 1.70/1.90 =========== start of search ===========
% 1.70/1.90
% 1.70/1.90 -------- PROOF --------
% 1.70/1.90
% 1.70/1.90 ----> UNIT CONFLICT at 0.01 sec ----> 224 [binary,223.1,185.1] $F.
% 1.70/1.90
% 1.70/1.90 Length of proof is 5. Level of proof is 3.
% 1.70/1.90
% 1.70/1.90 ---------------- PROOF ----------------
% 1.70/1.90 % SZS status Theorem
% 1.70/1.90 % SZS output start Refutation
% See solution above
% 1.70/1.90 ------------ end of proof -------------
% 1.70/1.90
% 1.70/1.90
% 1.70/1.90 Search stopped by max_proofs option.
% 1.70/1.90
% 1.70/1.90
% 1.70/1.90 Search stopped by max_proofs option.
% 1.70/1.90
% 1.70/1.90 ============ end of search ============
% 1.70/1.90
% 1.70/1.90 -------------- statistics -------------
% 1.70/1.90 clauses given 16
% 1.70/1.90 clauses generated 422
% 1.70/1.90 clauses kept 221
% 1.70/1.90 clauses forward subsumed 225
% 1.70/1.90 clauses back subsumed 2
% 1.70/1.90 Kbytes malloced 976
% 1.70/1.90
% 1.70/1.90 ----------- times (seconds) -----------
% 1.70/1.90 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.70/1.90 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.70/1.90 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.70/1.90
% 1.70/1.90 That finishes the proof of the theorem.
% 1.70/1.90
% 1.70/1.90 Process 25734 finished Wed Jul 27 10:44:07 2022
% 1.70/1.90 Otter interrupted
% 1.70/1.90 PROOF FOUND
%------------------------------------------------------------------------------