TSTP Solution File: SET146+3 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET146+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n023.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 13:13:11 EDT 2022
% Result : Theorem 1.93s 2.10s
% Output : Refutation 1.93s
% Verified :
% SZS Type : Refutation
% Derivation depth : 7
% Number of leaves : 8
% Syntax : Number of clauses : 17 ( 8 unt; 6 nHn; 9 RR)
% Number of literals : 27 ( 9 equ; 7 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 4 ( 2 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 2 con; 0-2 aty)
% Number of variables : 18 ( 4 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ member(A,intersection(B,C))
| member(A,B) ),
file('SET146+3.p',unknown),
[] ).
cnf(4,axiom,
~ member(A,empty_set),
file('SET146+3.p',unknown),
[] ).
cnf(10,axiom,
( ~ empty(A)
| ~ member(B,A) ),
file('SET146+3.p',unknown),
[] ).
cnf(14,axiom,
intersection(dollar_c1,empty_set) != empty_set,
file('SET146+3.p',unknown),
[] ).
cnf(18,axiom,
A = A,
file('SET146+3.p',unknown),
[] ).
cnf(19,axiom,
intersection(A,B) = intersection(B,A),
file('SET146+3.p',unknown),
[] ).
cnf(22,axiom,
( empty(A)
| member(dollar_f2(A),A) ),
file('SET146+3.p',unknown),
[] ).
cnf(23,axiom,
( A = B
| member(dollar_f3(A,B),A)
| member(dollar_f3(A,B),B) ),
file('SET146+3.p',unknown),
[] ).
cnf(29,plain,
intersection(empty_set,dollar_c1) != empty_set,
inference(para_from,[status(thm),theory(equality)],[19,14]),
[iquote('para_from,19.1.1,14.1.1')] ).
cnf(69,plain,
( empty_set = A
| member(dollar_f3(empty_set,A),A) ),
inference(hyper,[status(thm)],[23,4]),
[iquote('hyper,23,4')] ).
cnf(76,plain,
( A = empty_set
| member(dollar_f3(A,empty_set),A) ),
inference(hyper,[status(thm)],[23,4]),
[iquote('hyper,23,4')] ).
cnf(126,plain,
( empty_set = A
| member(dollar_f2(A),A) ),
inference(hyper,[status(thm)],[69,10,22]),
[iquote('hyper,69,10,22')] ).
cnf(157,plain,
( ~ member(A,B)
| member(dollar_f2(B),B) ),
inference(para_from,[status(thm),theory(equality)],[126,4]),
[iquote('para_from,126.1.1,4.1.2')] ).
cnf(162,plain,
( A = empty_set
| member(dollar_f2(A),A) ),
inference(hyper,[status(thm)],[76,157]),
[iquote('hyper,76,157')] ).
cnf(235,plain,
member(dollar_f2(intersection(empty_set,dollar_c1)),intersection(empty_set,dollar_c1)),
inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[162,29]),18]),
[iquote('para_from,162.1.1,29.1.1,unit_del,18')] ).
cnf(274,plain,
member(dollar_f2(intersection(empty_set,dollar_c1)),empty_set),
inference(hyper,[status(thm)],[235,1]),
[iquote('hyper,235,1')] ).
cnf(275,plain,
$false,
inference(binary,[status(thm)],[274,4]),
[iquote('binary,274.1,4.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SET146+3 : TPTP v8.1.0. Released v2.2.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.34 % Computer : n023.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Wed Jul 27 10:51:48 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.89/2.07 ----- Otter 3.3f, August 2004 -----
% 1.89/2.07 The process was started by sandbox2 on n023.cluster.edu,
% 1.89/2.07 Wed Jul 27 10:51:48 2022
% 1.89/2.07 The command was "./otter". The process ID is 28519.
% 1.89/2.07
% 1.89/2.07 set(prolog_style_variables).
% 1.89/2.07 set(auto).
% 1.89/2.07 dependent: set(auto1).
% 1.89/2.07 dependent: set(process_input).
% 1.89/2.07 dependent: clear(print_kept).
% 1.89/2.07 dependent: clear(print_new_demod).
% 1.89/2.07 dependent: clear(print_back_demod).
% 1.89/2.07 dependent: clear(print_back_sub).
% 1.89/2.07 dependent: set(control_memory).
% 1.89/2.07 dependent: assign(max_mem, 12000).
% 1.89/2.07 dependent: assign(pick_given_ratio, 4).
% 1.89/2.07 dependent: assign(stats_level, 1).
% 1.89/2.07 dependent: assign(max_seconds, 10800).
% 1.89/2.07 clear(print_given).
% 1.89/2.07
% 1.89/2.07 formula_list(usable).
% 1.89/2.07 all A (A=A).
% 1.89/2.07 all B C D (member(D,intersection(B,C))<->member(D,B)&member(D,C)).
% 1.89/2.07 all B (-member(B,empty_set)).
% 1.89/2.07 all B C (B=C<->subset(B,C)&subset(C,B)).
% 1.89/2.07 all B C (intersection(B,C)=intersection(C,B)).
% 1.89/2.07 all B C (subset(B,C)<-> (all D (member(D,B)->member(D,C)))).
% 1.89/2.07 all B subset(B,B).
% 1.89/2.07 all B (empty(B)<-> (all C (-member(C,B)))).
% 1.89/2.07 all B C (B=C<-> (all D (member(D,B)<->member(D,C)))).
% 1.89/2.07 -(all B (intersection(B,empty_set)=empty_set)).
% 1.89/2.07 end_of_list.
% 1.89/2.07
% 1.89/2.07 -------> usable clausifies to:
% 1.89/2.07
% 1.89/2.07 list(usable).
% 1.89/2.07 0 [] A=A.
% 1.89/2.07 0 [] -member(D,intersection(B,C))|member(D,B).
% 1.89/2.07 0 [] -member(D,intersection(B,C))|member(D,C).
% 1.89/2.07 0 [] member(D,intersection(B,C))| -member(D,B)| -member(D,C).
% 1.89/2.07 0 [] -member(B,empty_set).
% 1.89/2.07 0 [] B!=C|subset(B,C).
% 1.89/2.07 0 [] B!=C|subset(C,B).
% 1.89/2.07 0 [] B=C| -subset(B,C)| -subset(C,B).
% 1.89/2.07 0 [] intersection(B,C)=intersection(C,B).
% 1.89/2.07 0 [] -subset(B,C)| -member(D,B)|member(D,C).
% 1.89/2.07 0 [] subset(B,C)|member($f1(B,C),B).
% 1.89/2.07 0 [] subset(B,C)| -member($f1(B,C),C).
% 1.89/2.07 0 [] subset(B,B).
% 1.89/2.07 0 [] -empty(B)| -member(C,B).
% 1.89/2.07 0 [] empty(B)|member($f2(B),B).
% 1.89/2.07 0 [] B!=C| -member(D,B)|member(D,C).
% 1.89/2.07 0 [] B!=C|member(D,B)| -member(D,C).
% 1.89/2.07 0 [] B=C|member($f3(B,C),B)|member($f3(B,C),C).
% 1.89/2.07 0 [] B=C| -member($f3(B,C),B)| -member($f3(B,C),C).
% 1.89/2.07 0 [] intersection($c1,empty_set)!=empty_set.
% 1.89/2.07 end_of_list.
% 1.89/2.07
% 1.89/2.07 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.89/2.07
% 1.89/2.07 This ia a non-Horn set with equality. The strategy will be
% 1.89/2.07 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.89/2.07 deletion, with positive clauses in sos and nonpositive
% 1.89/2.07 clauses in usable.
% 1.89/2.07
% 1.89/2.07 dependent: set(knuth_bendix).
% 1.89/2.07 dependent: set(anl_eq).
% 1.89/2.07 dependent: set(para_from).
% 1.89/2.07 dependent: set(para_into).
% 1.89/2.07 dependent: clear(para_from_right).
% 1.89/2.07 dependent: clear(para_into_right).
% 1.89/2.07 dependent: set(para_from_vars).
% 1.89/2.07 dependent: set(eq_units_both_ways).
% 1.89/2.07 dependent: set(dynamic_demod_all).
% 1.89/2.07 dependent: set(dynamic_demod).
% 1.89/2.07 dependent: set(order_eq).
% 1.89/2.07 dependent: set(back_demod).
% 1.89/2.07 dependent: set(lrpo).
% 1.89/2.07 dependent: set(hyper_res).
% 1.89/2.07 dependent: set(unit_deletion).
% 1.89/2.07 dependent: set(factor).
% 1.89/2.07
% 1.89/2.07 ------------> process usable:
% 1.89/2.07 ** KEPT (pick-wt=8): 1 [] -member(A,intersection(B,C))|member(A,B).
% 1.89/2.07 ** KEPT (pick-wt=8): 2 [] -member(A,intersection(B,C))|member(A,C).
% 1.89/2.07 ** KEPT (pick-wt=11): 3 [] member(A,intersection(B,C))| -member(A,B)| -member(A,C).
% 1.89/2.07 ** KEPT (pick-wt=3): 4 [] -member(A,empty_set).
% 1.89/2.07 ** KEPT (pick-wt=6): 5 [] A!=B|subset(A,B).
% 1.89/2.07 ** KEPT (pick-wt=6): 6 [] A!=B|subset(B,A).
% 1.89/2.07 ** KEPT (pick-wt=9): 7 [] A=B| -subset(A,B)| -subset(B,A).
% 1.89/2.07 ** KEPT (pick-wt=9): 8 [] -subset(A,B)| -member(C,A)|member(C,B).
% 1.89/2.07 ** KEPT (pick-wt=8): 9 [] subset(A,B)| -member($f1(A,B),B).
% 1.89/2.07 ** KEPT (pick-wt=5): 10 [] -empty(A)| -member(B,A).
% 1.89/2.07 ** KEPT (pick-wt=9): 11 [] A!=B| -member(C,A)|member(C,B).
% 1.89/2.07 ** KEPT (pick-wt=9): 12 [] A!=B|member(C,A)| -member(C,B).
% 1.89/2.07 ** KEPT (pick-wt=13): 13 [] A=B| -member($f3(A,B),A)| -member($f3(A,B),B).
% 1.89/2.07 ** KEPT (pick-wt=5): 14 [] intersection($c1,empty_set)!=empty_set.
% 1.89/2.07
% 1.89/2.07 ------------> process sos:
% 1.89/2.07 ** KEPT (pick-wt=3): 18 [] A=A.
% 1.89/2.07 ** KEPT (pick-wt=7): 19 [] intersection(A,B)=intersection(B,A).
% 1.89/2.07 ** KEPT (pick-wt=8): 20 [] subset(A,B)|member($f1(A,B),A).
% 1.89/2.07 ** KEPT (pick-wt=3): 21 [] subset(A,A).
% 1.89/2.07 ** KEPT (pick-wt=6): 22 [] empty(A)|member($f2(A),A).
% 1.89/2.07 ** KEPT (pick-wt=13): 23 [] A=B|member($f3(A,B),A)|member($f3(A,B),B).
% 1.89/2.07 Following clause subsumed by 18 during input processing: 0 [copy,18,flip.1] A=A.
% 1.89/2.07 18 back subsumes 17.
% 1.89/2.07 18 back subsumes 16.
% 1.89/2.07 Following clause subsumed by 19 during input processing: 0 [copy,19,flip.1] intersection(A,B)=intersection(B,A).
% 1.93/2.10
% 1.93/2.10 ======= end of input processing =======
% 1.93/2.10
% 1.93/2.10 =========== start of search ===========
% 1.93/2.10
% 1.93/2.10 -------- PROOF --------
% 1.93/2.10
% 1.93/2.10 ----> UNIT CONFLICT at 0.02 sec ----> 275 [binary,274.1,4.1] $F.
% 1.93/2.10
% 1.93/2.10 Length of proof is 8. Level of proof is 6.
% 1.93/2.10
% 1.93/2.10 ---------------- PROOF ----------------
% 1.93/2.10 % SZS status Theorem
% 1.93/2.10 % SZS output start Refutation
% See solution above
% 1.93/2.10 ------------ end of proof -------------
% 1.93/2.10
% 1.93/2.10
% 1.93/2.10 Search stopped by max_proofs option.
% 1.93/2.10
% 1.93/2.10
% 1.93/2.10 Search stopped by max_proofs option.
% 1.93/2.10
% 1.93/2.10 ============ end of search ============
% 1.93/2.10
% 1.93/2.10 -------------- statistics -------------
% 1.93/2.10 clauses given 19
% 1.93/2.10 clauses generated 693
% 1.93/2.10 clauses kept 274
% 1.93/2.10 clauses forward subsumed 436
% 1.93/2.10 clauses back subsumed 2
% 1.93/2.10 Kbytes malloced 976
% 1.93/2.10
% 1.93/2.10 ----------- times (seconds) -----------
% 1.93/2.10 user CPU time 0.02 (0 hr, 0 min, 0 sec)
% 1.93/2.10 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.93/2.10 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.93/2.10
% 1.93/2.10 That finishes the proof of the theorem.
% 1.93/2.10
% 1.93/2.10 Process 28519 finished Wed Jul 27 10:51:50 2022
% 1.93/2.10 Otter interrupted
% 1.93/2.10 PROOF FOUND
%------------------------------------------------------------------------------