TSTP Solution File: SET603+3 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET603+3 : TPTP v8.1.0. Released v2.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-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 : 300s
% DateTime : Wed Jul 27 13:13:48 EDT 2022
% Result : Theorem 3.51s 3.75s
% Output : Refutation 3.51s
% Verified :
% SZS Type : Refutation
% Derivation depth : 6
% Number of leaves : 7
% Syntax : Number of clauses : 13 ( 7 unt; 4 nHn; 8 RR)
% Number of literals : 25 ( 7 equ; 6 neg)
% Maximal clause size : 4 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 2 con; 0-2 aty)
% Number of variables : 17 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( ~ member(dollar_f1(A,B),A)
| ~ member(dollar_f1(A,B),B)
| A = B ),
file('SET603+3.p',unknown),
[] ).
cnf(2,axiom,
( ~ member(A,difference(B,C))
| member(A,B) ),
file('SET603+3.p',unknown),
[] ).
cnf(4,axiom,
( member(A,difference(B,C))
| ~ member(A,B)
| member(A,C) ),
file('SET603+3.p',unknown),
[] ).
cnf(5,axiom,
~ member(A,empty_set),
file('SET603+3.p',unknown),
[] ).
cnf(15,axiom,
difference(dollar_c1,empty_set) != dollar_c1,
file('SET603+3.p',unknown),
[] ).
cnf(19,axiom,
A = A,
file('SET603+3.p',unknown),
[] ).
cnf(20,axiom,
( member(dollar_f1(A,B),A)
| member(dollar_f1(A,B),B)
| A = B ),
file('SET603+3.p',unknown),
[] ).
cnf(35,plain,
( member(dollar_f1(A,B),B)
| A = B
| member(dollar_f1(A,B),difference(A,C))
| member(dollar_f1(A,B),C) ),
inference(hyper,[status(thm)],[20,4]),
[iquote('hyper,20,4')] ).
cnf(44,plain,
( member(dollar_f1(A,difference(A,B)),difference(A,B))
| difference(A,B) = A
| member(dollar_f1(A,difference(A,B)),B) ),
inference(flip,[status(thm),theory(equality)],[inference(factor,[status(thm)],[35])]),
[iquote('factor,35.1.3,flip.2')] ).
cnf(3203,plain,
member(dollar_f1(dollar_c1,difference(dollar_c1,empty_set)),difference(dollar_c1,empty_set)),
inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[44,15]),19,5]),
[iquote('para_from,44.2.1,15.1.1,unit_del,19,5')] ).
cnf(3208,plain,
member(dollar_f1(dollar_c1,difference(dollar_c1,empty_set)),dollar_c1),
inference(hyper,[status(thm)],[3203,2]),
[iquote('hyper,3203,2')] ).
cnf(3210,plain,
difference(dollar_c1,empty_set) = dollar_c1,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[3208,1,3203])]),
[iquote('hyper,3208,1,3203,flip.1')] ).
cnf(3212,plain,
$false,
inference(binary,[status(thm)],[3210,15]),
[iquote('binary,3210.1,15.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11 % Problem : SET603+3 : TPTP v8.1.0. Released v2.2.0.
% 0.07/0.12 % Command : otter-tptp-script %s
% 0.13/0.33 % Computer : n005.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Wed Jul 27 10:40:50 EDT 2022
% 0.13/0.33 % CPUTime :
% 1.70/1.90 ----- Otter 3.3f, August 2004 -----
% 1.70/1.90 The process was started by sandbox2 on n005.cluster.edu,
% 1.70/1.90 Wed Jul 27 10:40:50 2022
% 1.70/1.90 The command was "./otter". The process ID is 22136.
% 1.70/1.90
% 1.70/1.90 set(prolog_style_variables).
% 1.70/1.90 set(auto).
% 1.70/1.90 dependent: set(auto1).
% 1.70/1.90 dependent: set(process_input).
% 1.70/1.90 dependent: clear(print_kept).
% 1.70/1.90 dependent: clear(print_new_demod).
% 1.70/1.90 dependent: clear(print_back_demod).
% 1.70/1.90 dependent: clear(print_back_sub).
% 1.70/1.90 dependent: set(control_memory).
% 1.70/1.90 dependent: assign(max_mem, 12000).
% 1.70/1.90 dependent: assign(pick_given_ratio, 4).
% 1.70/1.90 dependent: assign(stats_level, 1).
% 1.70/1.90 dependent: assign(max_seconds, 10800).
% 1.70/1.90 clear(print_given).
% 1.70/1.90
% 1.70/1.90 formula_list(usable).
% 1.70/1.90 all A (A=A).
% 1.70/1.90 all B C ((all D (member(D,B)<->member(D,C)))->B=C).
% 1.70/1.90 all B C D (member(D,difference(B,C))<->member(D,B)& -member(D,C)).
% 1.70/1.90 all B (-member(B,empty_set)).
% 1.70/1.90 all B C (B=C<->subset(B,C)&subset(C,B)).
% 1.70/1.90 all B C (B=C<-> (all D (member(D,B)<->member(D,C)))).
% 1.70/1.90 all B C (subset(B,C)<-> (all D (member(D,B)->member(D,C)))).
% 1.70/1.90 all B subset(B,B).
% 1.70/1.90 all B (empty(B)<-> (all C (-member(C,B)))).
% 1.70/1.90 -(all B (difference(B,empty_set)=B)).
% 1.70/1.90 end_of_list.
% 1.70/1.90
% 1.70/1.90 -------> usable clausifies to:
% 1.70/1.90
% 1.70/1.90 list(usable).
% 1.70/1.90 0 [] A=A.
% 1.70/1.90 0 [] member($f1(B,C),B)|member($f1(B,C),C)|B=C.
% 1.70/1.90 0 [] -member($f1(B,C),B)| -member($f1(B,C),C)|B=C.
% 1.70/1.90 0 [] -member(D,difference(B,C))|member(D,B).
% 1.70/1.90 0 [] -member(D,difference(B,C))| -member(D,C).
% 1.70/1.90 0 [] member(D,difference(B,C))| -member(D,B)|member(D,C).
% 1.70/1.90 0 [] -member(B,empty_set).
% 1.70/1.90 0 [] B!=C|subset(B,C).
% 1.70/1.90 0 [] B!=C|subset(C,B).
% 1.70/1.90 0 [] B=C| -subset(B,C)| -subset(C,B).
% 1.70/1.90 0 [] B!=C| -member(D,B)|member(D,C).
% 1.70/1.90 0 [] B!=C|member(D,B)| -member(D,C).
% 1.70/1.90 0 [] B=C|member($f2(B,C),B)|member($f2(B,C),C).
% 1.70/1.90 0 [] B=C| -member($f2(B,C),B)| -member($f2(B,C),C).
% 1.70/1.90 0 [] -subset(B,C)| -member(D,B)|member(D,C).
% 1.70/1.90 0 [] subset(B,C)|member($f3(B,C),B).
% 1.70/1.90 0 [] subset(B,C)| -member($f3(B,C),C).
% 1.70/1.90 0 [] subset(B,B).
% 1.70/1.90 0 [] -empty(B)| -member(C,B).
% 1.70/1.90 0 [] empty(B)|member($f4(B),B).
% 1.70/1.90 0 [] difference($c1,empty_set)!=$c1.
% 1.70/1.90 end_of_list.
% 1.70/1.90
% 1.70/1.90 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.70/1.90
% 1.70/1.90 This ia a non-Horn set with equality. The strategy will be
% 1.70/1.90 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.70/1.90 deletion, with positive clauses in sos and nonpositive
% 1.70/1.90 clauses in usable.
% 1.70/1.90
% 1.70/1.90 dependent: set(knuth_bendix).
% 1.70/1.90 dependent: set(anl_eq).
% 1.70/1.90 dependent: set(para_from).
% 1.70/1.90 dependent: set(para_into).
% 1.70/1.90 dependent: clear(para_from_right).
% 1.70/1.90 dependent: clear(para_into_right).
% 1.70/1.90 dependent: set(para_from_vars).
% 1.70/1.90 dependent: set(eq_units_both_ways).
% 1.70/1.90 dependent: set(dynamic_demod_all).
% 1.70/1.90 dependent: set(dynamic_demod).
% 1.70/1.90 dependent: set(order_eq).
% 1.70/1.90 dependent: set(back_demod).
% 1.70/1.90 dependent: set(lrpo).
% 1.70/1.90 dependent: set(hyper_res).
% 1.70/1.90 dependent: set(unit_deletion).
% 1.70/1.90 dependent: set(factor).
% 1.70/1.90
% 1.70/1.90 ------------> process usable:
% 1.70/1.90 ** KEPT (pick-wt=13): 1 [] -member($f1(A,B),A)| -member($f1(A,B),B)|A=B.
% 1.70/1.90 ** KEPT (pick-wt=8): 2 [] -member(A,difference(B,C))|member(A,B).
% 1.70/1.90 ** KEPT (pick-wt=8): 3 [] -member(A,difference(B,C))| -member(A,C).
% 1.70/1.90 ** KEPT (pick-wt=11): 4 [] member(A,difference(B,C))| -member(A,B)|member(A,C).
% 1.70/1.90 ** KEPT (pick-wt=3): 5 [] -member(A,empty_set).
% 1.70/1.90 ** KEPT (pick-wt=6): 6 [] A!=B|subset(A,B).
% 1.70/1.90 ** KEPT (pick-wt=6): 7 [] A!=B|subset(B,A).
% 1.70/1.90 ** KEPT (pick-wt=9): 8 [] A=B| -subset(A,B)| -subset(B,A).
% 1.70/1.90 ** KEPT (pick-wt=9): 9 [] A!=B| -member(C,A)|member(C,B).
% 1.70/1.90 ** KEPT (pick-wt=9): 10 [] A!=B|member(C,A)| -member(C,B).
% 1.70/1.90 ** KEPT (pick-wt=13): 11 [] A=B| -member($f2(A,B),A)| -member($f2(A,B),B).
% 1.70/1.90 ** KEPT (pick-wt=9): 12 [] -subset(A,B)| -member(C,A)|member(C,B).
% 1.70/1.90 ** KEPT (pick-wt=8): 13 [] subset(A,B)| -member($f3(A,B),B).
% 1.70/1.90 ** KEPT (pick-wt=5): 14 [] -empty(A)| -member(B,A).
% 1.70/1.90 ** KEPT (pick-wt=5): 15 [] difference($c1,empty_set)!=$c1.
% 1.70/1.90
% 1.70/1.90 ------------> process sos:
% 1.70/1.90 ** KEPT (pick-wt=3): 19 [] A=A.
% 1.70/1.90 ** KEPT (pick-wt=13): 20 [] member($f1(A,B),A)|member($f1(A,B),B)|A=B.
% 1.70/1.90 ** KEPT (pick-wt=13): 21 [] A=B|member($f2(A,B),A)|member($f2(A,B),B).
% 1.70/1.90 ** KEPT (pick-wt=8): 22 [] subset(A,B)|member($f3(A,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($f4(A),A).
% 1.70/1.90 Following clause subsumed by 19 during input processing: 0 [copy,19,flip.1] A=A.
% 1.70/1.90 19 back subsumes 18.
% 3.51/3.75 19 back subsumes 17.
% 3.51/3.75 19 back subsumes 16.
% 3.51/3.75
% 3.51/3.75 ======= end of input processing =======
% 3.51/3.75
% 3.51/3.75 =========== start of search ===========
% 3.51/3.75
% 3.51/3.75
% 3.51/3.75 Resetting weight limit to 12.
% 3.51/3.75
% 3.51/3.75
% 3.51/3.75 Resetting weight limit to 12.
% 3.51/3.75
% 3.51/3.75 sos_size=2173
% 3.51/3.75
% 3.51/3.75 -------- PROOF --------
% 3.51/3.75
% 3.51/3.75 ----> UNIT CONFLICT at 1.85 sec ----> 3212 [binary,3210.1,15.1] $F.
% 3.51/3.75
% 3.51/3.75 Length of proof is 5. Level of proof is 5.
% 3.51/3.75
% 3.51/3.75 ---------------- PROOF ----------------
% 3.51/3.75 % SZS status Theorem
% 3.51/3.75 % SZS output start Refutation
% See solution above
% 3.51/3.75 ------------ end of proof -------------
% 3.51/3.75
% 3.51/3.75
% 3.51/3.75 Search stopped by max_proofs option.
% 3.51/3.75
% 3.51/3.75
% 3.51/3.75 Search stopped by max_proofs option.
% 3.51/3.75
% 3.51/3.75 ============ end of search ============
% 3.51/3.75
% 3.51/3.75 -------------- statistics -------------
% 3.51/3.75 clauses given 102
% 3.51/3.75 clauses generated 15977
% 3.51/3.75 clauses kept 3208
% 3.51/3.75 clauses forward subsumed 6207
% 3.51/3.75 clauses back subsumed 305
% 3.51/3.75 Kbytes malloced 4882
% 3.51/3.75
% 3.51/3.75 ----------- times (seconds) -----------
% 3.51/3.75 user CPU time 1.85 (0 hr, 0 min, 1 sec)
% 3.51/3.75 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 3.51/3.75 wall-clock time 4 (0 hr, 0 min, 4 sec)
% 3.51/3.75
% 3.51/3.75 That finishes the proof of the theorem.
% 3.51/3.75
% 3.51/3.75 Process 22136 finished Wed Jul 27 10:40:54 2022
% 3.51/3.75 Otter interrupted
% 3.51/3.75 PROOF FOUND
%------------------------------------------------------------------------------