TSTP Solution File: SET929+1 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET929+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n019.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 13:14:33 EDT 2022
% Result : Theorem 1.75s 1.95s
% Output : Refutation 1.75s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 10
% Syntax : Number of clauses : 17 ( 6 unt; 4 nHn; 15 RR)
% Number of literals : 31 ( 8 equ; 12 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 : 6 ( 6 usr; 4 con; 0-2 aty)
% Number of variables : 16 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(3,axiom,
( set_difference(A,B) != empty_set
| subset(A,B) ),
file('SET929+1.p',unknown),
[] ).
cnf(4,axiom,
( set_difference(A,B) = empty_set
| ~ subset(A,B) ),
file('SET929+1.p',unknown),
[] ).
cnf(5,axiom,
( ~ subset(unordered_pair(A,B),C)
| in(A,C) ),
file('SET929+1.p',unknown),
[] ).
cnf(6,axiom,
( ~ subset(unordered_pair(A,B),C)
| in(B,C) ),
file('SET929+1.p',unknown),
[] ).
cnf(7,axiom,
( subset(unordered_pair(A,B),C)
| ~ in(A,C)
| ~ in(B,C) ),
file('SET929+1.p',unknown),
[] ).
cnf(8,axiom,
( set_difference(unordered_pair(dollar_c5,dollar_c4),dollar_c3) != empty_set
| ~ in(dollar_c5,dollar_c3)
| ~ in(dollar_c4,dollar_c3) ),
file('SET929+1.p',unknown),
[] ).
cnf(11,axiom,
A = A,
file('SET929+1.p',unknown),
[] ).
cnf(12,axiom,
unordered_pair(A,B) = unordered_pair(B,A),
file('SET929+1.p',unknown),
[] ).
cnf(16,axiom,
( set_difference(unordered_pair(dollar_c5,dollar_c4),dollar_c3) = empty_set
| in(dollar_c5,dollar_c3) ),
file('SET929+1.p',unknown),
[] ).
cnf(17,axiom,
( set_difference(unordered_pair(dollar_c5,dollar_c4),dollar_c3) = empty_set
| in(dollar_c4,dollar_c3) ),
file('SET929+1.p',unknown),
[] ).
cnf(23,plain,
( set_difference(unordered_pair(dollar_c4,dollar_c5),dollar_c3) != empty_set
| ~ in(dollar_c5,dollar_c3)
| ~ in(dollar_c4,dollar_c3) ),
inference(para_from,[status(thm),theory(equality)],[12,8]),
[iquote('para_from,12.1.1,8.1.1.1')] ).
cnf(32,plain,
( subset(unordered_pair(dollar_c5,dollar_c4),dollar_c3)
| in(dollar_c5,dollar_c3) ),
inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[16,3]),11]),
[iquote('para_from,16.1.1,3.1.1,unit_del,11')] ).
cnf(36,plain,
in(dollar_c5,dollar_c3),
inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[32,5])]),
[iquote('hyper,32,5,factor_simp')] ).
cnf(44,plain,
( subset(unordered_pair(dollar_c5,dollar_c4),dollar_c3)
| in(dollar_c4,dollar_c3) ),
inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[17,3]),11]),
[iquote('para_from,17.1.1,3.1.1,unit_del,11')] ).
cnf(45,plain,
in(dollar_c4,dollar_c3),
inference(factor_simp,[status(thm)],[inference(hyper,[status(thm)],[44,6])]),
[iquote('hyper,44,6,factor_simp')] ).
cnf(48,plain,
subset(unordered_pair(dollar_c4,dollar_c5),dollar_c3),
inference(hyper,[status(thm)],[45,7,36]),
[iquote('hyper,45,7,36')] ).
cnf(52,plain,
$false,
inference(unit_del,[status(thm)],[inference(para_into,[status(thm),theory(equality)],[23,4]),11,36,45,48]),
[iquote('para_into,23.1.1,4.1.1,unit_del,11,36,45,48')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : SET929+1 : TPTP v8.1.0. Released v3.2.0.
% 0.04/0.13 % Command : otter-tptp-script %s
% 0.13/0.34 % Computer : n019.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Wed Jul 27 10:31:07 EDT 2022
% 0.13/0.34 % CPUTime :
% 1.75/1.95 ----- Otter 3.3f, August 2004 -----
% 1.75/1.95 The process was started by sandbox2 on n019.cluster.edu,
% 1.75/1.95 Wed Jul 27 10:31:07 2022
% 1.75/1.95 The command was "./otter". The process ID is 23416.
% 1.75/1.95
% 1.75/1.95 set(prolog_style_variables).
% 1.75/1.95 set(auto).
% 1.75/1.95 dependent: set(auto1).
% 1.75/1.95 dependent: set(process_input).
% 1.75/1.95 dependent: clear(print_kept).
% 1.75/1.95 dependent: clear(print_new_demod).
% 1.75/1.95 dependent: clear(print_back_demod).
% 1.75/1.95 dependent: clear(print_back_sub).
% 1.75/1.95 dependent: set(control_memory).
% 1.75/1.95 dependent: assign(max_mem, 12000).
% 1.75/1.95 dependent: assign(pick_given_ratio, 4).
% 1.75/1.95 dependent: assign(stats_level, 1).
% 1.75/1.95 dependent: assign(max_seconds, 10800).
% 1.75/1.95 clear(print_given).
% 1.75/1.95
% 1.75/1.95 formula_list(usable).
% 1.75/1.95 all A (A=A).
% 1.75/1.95 all A B (in(A,B)-> -in(B,A)).
% 1.75/1.95 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.75/1.95 empty(empty_set).
% 1.75/1.95 exists A empty(A).
% 1.75/1.95 exists A (-empty(A)).
% 1.75/1.95 all A B subset(A,A).
% 1.75/1.95 all A B (set_difference(A,B)=empty_set<->subset(A,B)).
% 1.75/1.95 all A B C (subset(unordered_pair(A,B),C)<->in(A,C)&in(B,C)).
% 1.75/1.95 -(all A B C (set_difference(unordered_pair(A,B),C)=empty_set<->in(A,C)&in(B,C))).
% 1.75/1.95 end_of_list.
% 1.75/1.95
% 1.75/1.95 -------> usable clausifies to:
% 1.75/1.95
% 1.75/1.95 list(usable).
% 1.75/1.95 0 [] A=A.
% 1.75/1.95 0 [] -in(A,B)| -in(B,A).
% 1.75/1.95 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.95 0 [] empty(empty_set).
% 1.75/1.95 0 [] empty($c1).
% 1.75/1.95 0 [] -empty($c2).
% 1.75/1.95 0 [] subset(A,A).
% 1.75/1.95 0 [] set_difference(A,B)!=empty_set|subset(A,B).
% 1.75/1.95 0 [] set_difference(A,B)=empty_set| -subset(A,B).
% 1.75/1.95 0 [] -subset(unordered_pair(A,B),C)|in(A,C).
% 1.75/1.95 0 [] -subset(unordered_pair(A,B),C)|in(B,C).
% 1.75/1.95 0 [] subset(unordered_pair(A,B),C)| -in(A,C)| -in(B,C).
% 1.75/1.95 0 [] set_difference(unordered_pair($c5,$c4),$c3)=empty_set|in($c5,$c3).
% 1.75/1.95 0 [] set_difference(unordered_pair($c5,$c4),$c3)=empty_set|in($c4,$c3).
% 1.75/1.95 0 [] set_difference(unordered_pair($c5,$c4),$c3)!=empty_set| -in($c5,$c3)| -in($c4,$c3).
% 1.75/1.95 end_of_list.
% 1.75/1.95
% 1.75/1.95 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.75/1.95
% 1.75/1.95 This ia a non-Horn set with equality. The strategy will be
% 1.75/1.95 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.75/1.95 deletion, with positive clauses in sos and nonpositive
% 1.75/1.95 clauses in usable.
% 1.75/1.95
% 1.75/1.95 dependent: set(knuth_bendix).
% 1.75/1.95 dependent: set(anl_eq).
% 1.75/1.95 dependent: set(para_from).
% 1.75/1.95 dependent: set(para_into).
% 1.75/1.95 dependent: clear(para_from_right).
% 1.75/1.95 dependent: clear(para_into_right).
% 1.75/1.95 dependent: set(para_from_vars).
% 1.75/1.95 dependent: set(eq_units_both_ways).
% 1.75/1.95 dependent: set(dynamic_demod_all).
% 1.75/1.95 dependent: set(dynamic_demod).
% 1.75/1.95 dependent: set(order_eq).
% 1.75/1.95 dependent: set(back_demod).
% 1.75/1.95 dependent: set(lrpo).
% 1.75/1.95 dependent: set(hyper_res).
% 1.75/1.95 dependent: set(unit_deletion).
% 1.75/1.95 dependent: set(factor).
% 1.75/1.95
% 1.75/1.95 ------------> process usable:
% 1.75/1.95 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.75/1.95 ** KEPT (pick-wt=2): 2 [] -empty($c2).
% 1.75/1.95 ** KEPT (pick-wt=8): 3 [] set_difference(A,B)!=empty_set|subset(A,B).
% 1.75/1.95 ** KEPT (pick-wt=8): 4 [] set_difference(A,B)=empty_set| -subset(A,B).
% 1.75/1.95 ** KEPT (pick-wt=8): 5 [] -subset(unordered_pair(A,B),C)|in(A,C).
% 1.75/1.95 ** KEPT (pick-wt=8): 6 [] -subset(unordered_pair(A,B),C)|in(B,C).
% 1.75/1.95 ** KEPT (pick-wt=11): 7 [] subset(unordered_pair(A,B),C)| -in(A,C)| -in(B,C).
% 1.75/1.95 ** KEPT (pick-wt=13): 8 [] set_difference(unordered_pair($c5,$c4),$c3)!=empty_set| -in($c5,$c3)| -in($c4,$c3).
% 1.75/1.95
% 1.75/1.95 ------------> process sos:
% 1.75/1.95 ** KEPT (pick-wt=3): 11 [] A=A.
% 1.75/1.95 ** KEPT (pick-wt=7): 12 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.95 ** KEPT (pick-wt=2): 13 [] empty(empty_set).
% 1.75/1.95 ** KEPT (pick-wt=2): 14 [] empty($c1).
% 1.75/1.95 ** KEPT (pick-wt=3): 15 [] subset(A,A).
% 1.75/1.95 ** KEPT (pick-wt=10): 16 [] set_difference(unordered_pair($c5,$c4),$c3)=empty_set|in($c5,$c3).
% 1.75/1.95 ** KEPT (pick-wt=10): 17 [] set_difference(unordered_pair($c5,$c4),$c3)=empty_set|in($c4,$c3).
% 1.75/1.95 Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.75/1.95 Following clause subsumed by 12 during input processing: 0 [copy,12,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.95
% 1.75/1.95 ======= end of input processing =======
% 1.75/1.95
% 1.75/1.95 =========== start of search ===========
% 1.75/1.95
% 1.75/1.95 -------- PROOF --------
% 1.75/1.95
% 1.75/1.95 -----> EMPTY CLAUSE at 0.00 sec ----> 52 [para_into,23.1.1,4.1.1,unit_del,11,36,45,48] $F.
% 1.75/1.95
% 1.75/1.95 Length of proof is 6. Level of proof is 3.
% 1.75/1.95
% 1.75/1.95 ---------------- PROOF ----------------
% 1.75/1.95 % SZS status Theorem
% 1.75/1.95 % SZS output start Refutation
% See solution above
% 1.75/1.95 ------------ end of proof -------------
% 1.75/1.95
% 1.75/1.95
% 1.75/1.95 Search stopped by max_proofs option.
% 1.75/1.95
% 1.75/1.95
% 1.75/1.95 Search stopped by max_proofs option.
% 1.75/1.95
% 1.75/1.95 ============ end of search ============
% 1.75/1.95
% 1.75/1.95 -------------- statistics -------------
% 1.75/1.95 clauses given 21
% 1.75/1.95 clauses generated 96
% 1.75/1.95 clauses kept 45
% 1.75/1.95 clauses forward subsumed 65
% 1.75/1.95 clauses back subsumed 11
% 1.75/1.95 Kbytes malloced 976
% 1.75/1.95
% 1.75/1.95 ----------- times (seconds) -----------
% 1.75/1.95 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.95 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.95 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.75/1.95
% 1.75/1.95 That finishes the proof of the theorem.
% 1.75/1.95
% 1.75/1.95 Process 23416 finished Wed Jul 27 10:31:08 2022
% 1.75/1.95 Otter interrupted
% 1.75/1.95 PROOF FOUND
%------------------------------------------------------------------------------