TSTP Solution File: SEU123+1 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU123+1 : TPTP v8.1.0. Released v3.3.0.
% 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 13:14:49 EDT 2022
% Result : Theorem 1.68s 1.89s
% Output : Refutation 1.68s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 7
% Syntax : Number of clauses : 14 ( 12 unt; 0 nHn; 12 RR)
% Number of literals : 18 ( 7 equ; 7 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 1 ( 1 avg)
% Number of predicates : 4 ( 2 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 3 con; 0-0 aty)
% Number of variables : 6 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(4,axiom,
( A = B
| ~ subset(A,B)
| ~ subset(B,A) ),
file('SEU123+1.p',unknown),
[] ).
cnf(6,axiom,
dollar_c3 != empty_set,
file('SEU123+1.p',unknown),
[] ).
cnf(7,plain,
empty_set != dollar_c3,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[6])]),
[iquote('copy,6,flip.1')] ).
cnf(10,axiom,
( ~ empty(A)
| A = B
| ~ empty(B) ),
file('SEU123+1.p',unknown),
[] ).
cnf(12,axiom,
empty(empty_set),
file('SEU123+1.p',unknown),
[] ).
cnf(13,axiom,
empty(dollar_c1),
file('SEU123+1.p',unknown),
[] ).
cnf(15,axiom,
subset(empty_set,A),
file('SEU123+1.p',unknown),
[] ).
cnf(16,axiom,
subset(dollar_c3,empty_set),
file('SEU123+1.p',unknown),
[] ).
cnf(19,plain,
empty_set = dollar_c1,
inference(hyper,[status(thm)],[13,10,12]),
[iquote('hyper,13,10,12')] ).
cnf(21,plain,
subset(dollar_c3,dollar_c1),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[16]),19]),
[iquote('back_demod,16,demod,19')] ).
cnf(22,plain,
subset(dollar_c1,A),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[15]),19]),
[iquote('back_demod,15,demod,19')] ).
cnf(24,plain,
dollar_c3 != dollar_c1,
inference(flip,[status(thm),theory(equality)],[inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[7]),19])]),
[iquote('back_demod,7,demod,19,flip.1')] ).
cnf(34,plain,
dollar_c3 = dollar_c1,
inference(hyper,[status(thm)],[22,4,21]),
[iquote('hyper,22,4,21')] ).
cnf(36,plain,
$false,
inference(binary,[status(thm)],[34,24]),
[iquote('binary,34.1,24.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SEU123+1 : TPTP v8.1.0. Released v3.3.0.
% 0.07/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n005.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 07:58:05 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.68/1.89 ----- Otter 3.3f, August 2004 -----
% 1.68/1.89 The process was started by sandbox2 on n005.cluster.edu,
% 1.68/1.89 Wed Jul 27 07:58:05 2022
% 1.68/1.89 The command was "./otter". The process ID is 18169.
% 1.68/1.89
% 1.68/1.89 set(prolog_style_variables).
% 1.68/1.89 set(auto).
% 1.68/1.89 dependent: set(auto1).
% 1.68/1.89 dependent: set(process_input).
% 1.68/1.89 dependent: clear(print_kept).
% 1.68/1.89 dependent: clear(print_new_demod).
% 1.68/1.89 dependent: clear(print_back_demod).
% 1.68/1.89 dependent: clear(print_back_sub).
% 1.68/1.89 dependent: set(control_memory).
% 1.68/1.89 dependent: assign(max_mem, 12000).
% 1.68/1.89 dependent: assign(pick_given_ratio, 4).
% 1.68/1.89 dependent: assign(stats_level, 1).
% 1.68/1.89 dependent: assign(max_seconds, 10800).
% 1.68/1.89 clear(print_given).
% 1.68/1.89
% 1.68/1.89 formula_list(usable).
% 1.68/1.89 all A (A=A).
% 1.68/1.89 all A B (in(A,B)-> -in(B,A)).
% 1.68/1.89 all A B (A=B<->subset(A,B)&subset(B,A)).
% 1.68/1.89 $T.
% 1.68/1.89 empty(empty_set).
% 1.68/1.89 exists A empty(A).
% 1.68/1.89 exists A (-empty(A)).
% 1.68/1.89 all A B subset(A,A).
% 1.68/1.89 all A subset(empty_set,A).
% 1.68/1.89 -(all A (subset(A,empty_set)->A=empty_set)).
% 1.68/1.89 all A (empty(A)->A=empty_set).
% 1.68/1.89 all A B (-(in(A,B)&empty(B))).
% 1.68/1.89 all A B (-(empty(A)&A!=B&empty(B))).
% 1.68/1.89 end_of_list.
% 1.68/1.89
% 1.68/1.89 -------> usable clausifies to:
% 1.68/1.89
% 1.68/1.89 list(usable).
% 1.68/1.89 0 [] A=A.
% 1.68/1.89 0 [] -in(A,B)| -in(B,A).
% 1.68/1.89 0 [] A!=B|subset(A,B).
% 1.68/1.89 0 [] A!=B|subset(B,A).
% 1.68/1.89 0 [] A=B| -subset(A,B)| -subset(B,A).
% 1.68/1.89 0 [] $T.
% 1.68/1.89 0 [] empty(empty_set).
% 1.68/1.89 0 [] empty($c1).
% 1.68/1.89 0 [] -empty($c2).
% 1.68/1.89 0 [] subset(A,A).
% 1.68/1.89 0 [] subset(empty_set,A).
% 1.68/1.89 0 [] subset($c3,empty_set).
% 1.68/1.89 0 [] $c3!=empty_set.
% 1.68/1.89 0 [] -empty(A)|A=empty_set.
% 1.68/1.89 0 [] -in(A,B)| -empty(B).
% 1.68/1.89 0 [] -empty(A)|A=B| -empty(B).
% 1.68/1.89 end_of_list.
% 1.68/1.89
% 1.68/1.89 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.68/1.89
% 1.68/1.89 This is a Horn set with equality. The strategy will be
% 1.68/1.89 Knuth-Bendix and hyper_res, with positive clauses in
% 1.68/1.89 sos and nonpositive clauses in usable.
% 1.68/1.89
% 1.68/1.89 dependent: set(knuth_bendix).
% 1.68/1.89 dependent: set(anl_eq).
% 1.68/1.89 dependent: set(para_from).
% 1.68/1.89 dependent: set(para_into).
% 1.68/1.89 dependent: clear(para_from_right).
% 1.68/1.89 dependent: clear(para_into_right).
% 1.68/1.89 dependent: set(para_from_vars).
% 1.68/1.89 dependent: set(eq_units_both_ways).
% 1.68/1.89 dependent: set(dynamic_demod_all).
% 1.68/1.89 dependent: set(dynamic_demod).
% 1.68/1.89 dependent: set(order_eq).
% 1.68/1.89 dependent: set(back_demod).
% 1.68/1.89 dependent: set(lrpo).
% 1.68/1.89 dependent: set(hyper_res).
% 1.68/1.89 dependent: clear(order_hyper).
% 1.68/1.89
% 1.68/1.89 ------------> process usable:
% 1.68/1.89 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.68/1.89 ** KEPT (pick-wt=6): 2 [] A!=B|subset(A,B).
% 1.68/1.89 ** KEPT (pick-wt=6): 3 [] A!=B|subset(B,A).
% 1.68/1.89 ** KEPT (pick-wt=9): 4 [] A=B| -subset(A,B)| -subset(B,A).
% 1.68/1.89 ** KEPT (pick-wt=2): 5 [] -empty($c2).
% 1.68/1.89 ** KEPT (pick-wt=3): 7 [copy,6,flip.1] empty_set!=$c3.
% 1.68/1.89 ** KEPT (pick-wt=5): 8 [] -empty(A)|A=empty_set.
% 1.68/1.89 ** KEPT (pick-wt=5): 9 [] -in(A,B)| -empty(B).
% 1.68/1.89 ** KEPT (pick-wt=7): 10 [] -empty(A)|A=B| -empty(B).
% 1.68/1.89
% 1.68/1.89 ------------> process sos:
% 1.68/1.89 ** KEPT (pick-wt=3): 11 [] A=A.
% 1.68/1.89 ** KEPT (pick-wt=2): 12 [] empty(empty_set).
% 1.68/1.89 ** KEPT (pick-wt=2): 13 [] empty($c1).
% 1.68/1.89 ** KEPT (pick-wt=3): 14 [] subset(A,A).
% 1.68/1.89 ** KEPT (pick-wt=3): 15 [] subset(empty_set,A).
% 1.68/1.89 ** KEPT (pick-wt=3): 16 [] subset($c3,empty_set).
% 1.68/1.89 Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.68/1.89
% 1.68/1.89 ======= end of input processing =======
% 1.68/1.89
% 1.68/1.89 =========== start of search ===========
% 1.68/1.89
% 1.68/1.89 -------- PROOF --------
% 1.68/1.89
% 1.68/1.89 ----> UNIT CONFLICT at 0.00 sec ----> 36 [binary,34.1,24.1] $F.
% 1.68/1.89
% 1.68/1.89 Length of proof is 6. Level of proof is 3.
% 1.68/1.89
% 1.68/1.89 ---------------- PROOF ----------------
% 1.68/1.89 % SZS status Theorem
% 1.68/1.89 % SZS output start Refutation
% See solution above
% 1.68/1.89 ------------ end of proof -------------
% 1.68/1.89
% 1.68/1.89
% 1.68/1.89 Search stopped by max_proofs option.
% 1.68/1.89
% 1.68/1.89
% 1.68/1.89 Search stopped by max_proofs option.
% 1.68/1.89
% 1.68/1.89 ============ end of search ============
% 1.68/1.89
% 1.68/1.89 -------------- statistics -------------
% 1.68/1.89 clauses given 8
% 1.68/1.89 clauses generated 39
% 1.68/1.89 clauses kept 32
% 1.68/1.89 clauses forward subsumed 26
% 1.68/1.89 clauses back subsumed 0
% 1.68/1.89 Kbytes malloced 976
% 1.68/1.89
% 1.68/1.89 ----------- times (seconds) -----------
% 1.68/1.89 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.89 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.68/1.89 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.68/1.89
% 1.68/1.89 That finishes the proof of the theorem.
% 1.68/1.89
% 1.68/1.89 Process 18169 finished Wed Jul 27 07:58:07 2022
% 1.68/1.89 Otter interrupted
% 1.68/1.89 PROOF FOUND
%------------------------------------------------------------------------------