TSTP Solution File: SET903+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET903+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n028.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 13:14:29 EDT 2022
% Result : Theorem 1.66s 1.90s
% Output : Refutation 1.66s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 6
% Syntax : Number of clauses : 10 ( 8 unt; 2 nHn; 10 RR)
% Number of literals : 14 ( 13 equ; 7 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 4 con; 0-2 aty)
% Number of variables : 6 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(4,axiom,
( singleton(A) != set_union2(B,C)
| B = singleton(A)
| B = empty_set ),
file('SET903+1.p',unknown),
[] ).
cnf(6,axiom,
( singleton(A) != set_union2(B,C)
| C = singleton(A)
| C = empty_set ),
file('SET903+1.p',unknown),
[] ).
cnf(7,axiom,
dollar_c4 != dollar_c3,
file('SET903+1.p',unknown),
[] ).
cnf(8,axiom,
dollar_c4 != empty_set,
file('SET903+1.p',unknown),
[] ).
cnf(9,plain,
empty_set != dollar_c4,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[8])]),
[iquote('copy,8,flip.1')] ).
cnf(10,axiom,
dollar_c3 != empty_set,
file('SET903+1.p',unknown),
[] ).
cnf(11,plain,
empty_set != dollar_c3,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[10])]),
[iquote('copy,10,flip.1')] ).
cnf(20,axiom,
singleton(dollar_c5) = set_union2(dollar_c4,dollar_c3),
file('SET903+1.p',unknown),
[] ).
cnf(35,plain,
set_union2(dollar_c4,dollar_c3) = dollar_c3,
inference(flip,[status(thm),theory(equality)],[inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[20,6]),20]),11])]),
[iquote('hyper,19,6,demod,20,unit_del,11,flip.1')] ).
cnf(36,plain,
$false,
inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[20,4]),20,35]),7,9]),
[iquote('hyper,19,4,demod,20,35,unit_del,7,9')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.12 % Problem : SET903+1 : TPTP v8.1.0. Released v3.2.0.
% 0.03/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n028.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 10:33:49 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.66/1.90 ----- Otter 3.3f, August 2004 -----
% 1.66/1.90 The process was started by sandbox on n028.cluster.edu,
% 1.66/1.90 Wed Jul 27 10:33:49 2022
% 1.66/1.90 The command was "./otter". The process ID is 8925.
% 1.66/1.90
% 1.66/1.90 set(prolog_style_variables).
% 1.66/1.90 set(auto).
% 1.66/1.90 dependent: set(auto1).
% 1.66/1.90 dependent: set(process_input).
% 1.66/1.90 dependent: clear(print_kept).
% 1.66/1.90 dependent: clear(print_new_demod).
% 1.66/1.90 dependent: clear(print_back_demod).
% 1.66/1.90 dependent: clear(print_back_sub).
% 1.66/1.90 dependent: set(control_memory).
% 1.66/1.90 dependent: assign(max_mem, 12000).
% 1.66/1.90 dependent: assign(pick_given_ratio, 4).
% 1.66/1.90 dependent: assign(stats_level, 1).
% 1.66/1.90 dependent: assign(max_seconds, 10800).
% 1.66/1.90 clear(print_given).
% 1.66/1.90
% 1.66/1.90 formula_list(usable).
% 1.66/1.90 all A (A=A).
% 1.66/1.90 all A B (set_union2(A,B)=set_union2(B,A)).
% 1.66/1.90 empty(empty_set).
% 1.66/1.90 all A B (-empty(A)-> -empty(set_union2(A,B))).
% 1.66/1.90 all A B (-empty(A)-> -empty(set_union2(B,A))).
% 1.66/1.90 all A B (set_union2(A,A)=A).
% 1.66/1.90 exists A empty(A).
% 1.66/1.90 exists A (-empty(A)).
% 1.66/1.90 all A B C (-(singleton(A)=set_union2(B,C)& -(B=singleton(A)&C=singleton(A))& -(B=empty_set&C=singleton(A))& -(B=singleton(A)&C=empty_set))).
% 1.66/1.90 -(all A B C (-(singleton(A)=set_union2(B,C)&B!=C&B!=empty_set&C!=empty_set))).
% 1.66/1.90 end_of_list.
% 1.66/1.90
% 1.66/1.90 -------> usable clausifies to:
% 1.66/1.90
% 1.66/1.90 list(usable).
% 1.66/1.90 0 [] A=A.
% 1.66/1.90 0 [] set_union2(A,B)=set_union2(B,A).
% 1.66/1.90 0 [] empty(empty_set).
% 1.66/1.90 0 [] empty(A)| -empty(set_union2(A,B)).
% 1.66/1.90 0 [] empty(A)| -empty(set_union2(B,A)).
% 1.66/1.90 0 [] set_union2(A,A)=A.
% 1.66/1.90 0 [] empty($c1).
% 1.66/1.90 0 [] -empty($c2).
% 1.66/1.90 0 [] singleton(A)!=set_union2(B,C)|B=singleton(A)|B=empty_set.
% 1.66/1.90 0 [] singleton(A)!=set_union2(B,C)|B=singleton(A)|C=singleton(A).
% 1.66/1.90 0 [] singleton(A)!=set_union2(B,C)|C=singleton(A)|C=empty_set.
% 1.66/1.90 0 [] singleton($c5)=set_union2($c4,$c3).
% 1.66/1.90 0 [] $c4!=$c3.
% 1.66/1.90 0 [] $c4!=empty_set.
% 1.66/1.90 0 [] $c3!=empty_set.
% 1.66/1.90 end_of_list.
% 1.66/1.90
% 1.66/1.90 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.66/1.90
% 1.66/1.90 This ia a non-Horn set with equality. The strategy will be
% 1.66/1.90 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.66/1.90 deletion, with positive clauses in sos and nonpositive
% 1.66/1.90 clauses in usable.
% 1.66/1.90
% 1.66/1.90 dependent: set(knuth_bendix).
% 1.66/1.90 dependent: set(anl_eq).
% 1.66/1.90 dependent: set(para_from).
% 1.66/1.90 dependent: set(para_into).
% 1.66/1.90 dependent: clear(para_from_right).
% 1.66/1.90 dependent: clear(para_into_right).
% 1.66/1.90 dependent: set(para_from_vars).
% 1.66/1.90 dependent: set(eq_units_both_ways).
% 1.66/1.90 dependent: set(dynamic_demod_all).
% 1.66/1.90 dependent: set(dynamic_demod).
% 1.66/1.90 dependent: set(order_eq).
% 1.66/1.90 dependent: set(back_demod).
% 1.66/1.90 dependent: set(lrpo).
% 1.66/1.90 dependent: set(hyper_res).
% 1.66/1.90 dependent: set(unit_deletion).
% 1.66/1.90 dependent: set(factor).
% 1.66/1.90
% 1.66/1.90 ------------> process usable:
% 1.66/1.90 ** KEPT (pick-wt=6): 1 [] empty(A)| -empty(set_union2(A,B)).
% 1.66/1.90 ** KEPT (pick-wt=6): 2 [] empty(A)| -empty(set_union2(B,A)).
% 1.66/1.90 ** KEPT (pick-wt=2): 3 [] -empty($c2).
% 1.66/1.90 ** KEPT (pick-wt=13): 4 [] singleton(A)!=set_union2(B,C)|B=singleton(A)|B=empty_set.
% 1.66/1.90 ** KEPT (pick-wt=14): 5 [] singleton(A)!=set_union2(B,C)|B=singleton(A)|C=singleton(A).
% 1.66/1.90 ** KEPT (pick-wt=13): 6 [] singleton(A)!=set_union2(B,C)|C=singleton(A)|C=empty_set.
% 1.66/1.90 ** KEPT (pick-wt=3): 7 [] $c4!=$c3.
% 1.66/1.90 ** KEPT (pick-wt=3): 9 [copy,8,flip.1] empty_set!=$c4.
% 1.66/1.90 ** KEPT (pick-wt=3): 11 [copy,10,flip.1] empty_set!=$c3.
% 1.66/1.90
% 1.66/1.90 ------------> process sos:
% 1.66/1.90 ** KEPT (pick-wt=3): 13 [] A=A.
% 1.66/1.90 ** KEPT (pick-wt=7): 14 [] set_union2(A,B)=set_union2(B,A).
% 1.66/1.90 ** KEPT (pick-wt=2): 15 [] empty(empty_set).
% 1.66/1.90 ** KEPT (pick-wt=5): 16 [] set_union2(A,A)=A.
% 1.66/1.90 ---> New Demodulator: 17 [new_demod,16] set_union2(A,A)=A.
% 1.66/1.90 ** KEPT (pick-wt=2): 18 [] empty($c1).
% 1.66/1.90 ** KEPT (pick-wt=6): 19 [] singleton($c5)=set_union2($c4,$c3).
% 1.66/1.90 ---> New Demodulator: 20 [new_demod,19] singleton($c5)=set_union2($c4,$c3).
% 1.66/1.90 Following clause subsumed by 13 during input processing: 0 [copy,13,flip.1] A=A.
% 1.66/1.90 Following clause subsumed by 14 during input processing: 0 [copy,14,flip.1] set_union2(A,B)=set_union2(B,A).
% 1.66/1.90 >>>> Starting back demodulation with 17.
% 1.66/1.90 >> back demodulating 12 with 17.
% 1.66/1.90 >>>> Starting back demodulation with 20.
% 1.66/1.90
% 1.66/1.90 ======= end of input processing =======
% 1.66/1.90
% 1.66/1.90 =========== start of search ===========
% 1.66/1.90
% 1.66/1.90 -------- PROOF --------
% 1.66/1.90
% 1.66/1.90 -----> EMPTY CLAUSE at 0.00 sec ----> 36 [hyper,19,4,demod,20,35,unit_del,7,9] $F.
% 1.66/1.90
% 1.66/1.90 Length of proof is 3. Level of proof is 2.
% 1.66/1.90
% 1.66/1.90 ---------------- PROOF ----------------
% 1.66/1.90 % SZS status Theorem
% 1.66/1.90 % SZS output start Refutation
% See solution above
% 1.66/1.90 ------------ end of proof -------------
% 1.66/1.90
% 1.66/1.90
% 1.66/1.90 Search stopped by max_proofs option.
% 1.66/1.90
% 1.66/1.90
% 1.66/1.90 Search stopped by max_proofs option.
% 1.66/1.90
% 1.66/1.90 ============ end of search ============
% 1.66/1.90
% 1.66/1.90 -------------- statistics -------------
% 1.66/1.90 clauses given 5
% 1.66/1.90 clauses generated 28
% 1.66/1.90 clauses kept 30
% 1.66/1.90 clauses forward subsumed 13
% 1.66/1.90 clauses back subsumed 0
% 1.66/1.90 Kbytes malloced 976
% 1.66/1.90
% 1.66/1.90 ----------- times (seconds) -----------
% 1.66/1.90 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.90 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.90 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.66/1.90
% 1.66/1.90 That finishes the proof of the theorem.
% 1.66/1.90
% 1.66/1.90 Process 8925 finished Wed Jul 27 10:33:50 2022
% 1.66/1.90 Otter interrupted
% 1.66/1.90 PROOF FOUND
%------------------------------------------------------------------------------