TSTP Solution File: SET881+1 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET881+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n004.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 13:14:27 EDT 2022
% Result : Theorem 1.70s 1.89s
% Output : Refutation 1.70s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 4
% Syntax : Number of clauses : 7 ( 5 unt; 0 nHn; 4 RR)
% Number of literals : 10 ( 6 equ; 4 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 3 con; 0-2 aty)
% Number of variables : 11 ( 3 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(3,axiom,
( A != unordered_pair(B,C)
| in(D,A)
| D != B ),
file('SET881+1.p',unknown),
[] ).
cnf(8,axiom,
( set_difference(singleton(A),B) = empty_set
| ~ in(A,B) ),
file('SET881+1.p',unknown),
[] ).
cnf(10,axiom,
set_difference(singleton(dollar_c4),unordered_pair(dollar_c4,dollar_c3)) != empty_set,
file('SET881+1.p',unknown),
[] ).
cnf(13,axiom,
A = A,
file('SET881+1.p',unknown),
[] ).
cnf(20,plain,
in(A,unordered_pair(A,B)),
inference(hyper,[status(thm)],[13,3,13]),
[iquote('hyper,13,3,13')] ).
cnf(38,plain,
set_difference(singleton(A),unordered_pair(A,B)) = empty_set,
inference(hyper,[status(thm)],[20,8]),
[iquote('hyper,20,8')] ).
cnf(40,plain,
$false,
inference(binary,[status(thm)],[38,10]),
[iquote('binary,38.1,10.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SET881+1 : TPTP v8.1.0. Released v3.2.0.
% 0.11/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n004.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 10:35:36 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.70/1.89 ----- Otter 3.3f, August 2004 -----
% 1.70/1.89 The process was started by sandbox2 on n004.cluster.edu,
% 1.70/1.89 Wed Jul 27 10:35:36 2022
% 1.70/1.89 The command was "./otter". The process ID is 14993.
% 1.70/1.89
% 1.70/1.89 set(prolog_style_variables).
% 1.70/1.89 set(auto).
% 1.70/1.89 dependent: set(auto1).
% 1.70/1.89 dependent: set(process_input).
% 1.70/1.89 dependent: clear(print_kept).
% 1.70/1.89 dependent: clear(print_new_demod).
% 1.70/1.89 dependent: clear(print_back_demod).
% 1.70/1.89 dependent: clear(print_back_sub).
% 1.70/1.89 dependent: set(control_memory).
% 1.70/1.89 dependent: assign(max_mem, 12000).
% 1.70/1.89 dependent: assign(pick_given_ratio, 4).
% 1.70/1.89 dependent: assign(stats_level, 1).
% 1.70/1.89 dependent: assign(max_seconds, 10800).
% 1.70/1.89 clear(print_given).
% 1.70/1.89
% 1.70/1.89 formula_list(usable).
% 1.70/1.89 all A (A=A).
% 1.70/1.89 all A B (in(A,B)-> -in(B,A)).
% 1.70/1.89 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.70/1.89 all A B C (C=unordered_pair(A,B)<-> (all D (in(D,C)<->D=A|D=B))).
% 1.70/1.89 empty(empty_set).
% 1.70/1.89 all A B (set_difference(singleton(A),B)=empty_set<->in(A,B)).
% 1.70/1.89 exists A empty(A).
% 1.70/1.89 exists A (-empty(A)).
% 1.70/1.89 -(all A B (set_difference(singleton(A),unordered_pair(A,B))=empty_set)).
% 1.70/1.89 end_of_list.
% 1.70/1.89
% 1.70/1.89 -------> usable clausifies to:
% 1.70/1.89
% 1.70/1.89 list(usable).
% 1.70/1.89 0 [] A=A.
% 1.70/1.89 0 [] -in(A,B)| -in(B,A).
% 1.70/1.89 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.70/1.89 0 [] C!=unordered_pair(A,B)| -in(D,C)|D=A|D=B.
% 1.70/1.89 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=A.
% 1.70/1.89 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=B.
% 1.70/1.89 0 [] C=unordered_pair(A,B)|in($f1(A,B,C),C)|$f1(A,B,C)=A|$f1(A,B,C)=B.
% 1.70/1.89 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=A.
% 1.70/1.89 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=B.
% 1.70/1.89 0 [] empty(empty_set).
% 1.70/1.89 0 [] set_difference(singleton(A),B)!=empty_set|in(A,B).
% 1.70/1.89 0 [] set_difference(singleton(A),B)=empty_set| -in(A,B).
% 1.70/1.89 0 [] empty($c1).
% 1.70/1.89 0 [] -empty($c2).
% 1.70/1.89 0 [] set_difference(singleton($c4),unordered_pair($c4,$c3))!=empty_set.
% 1.70/1.89 end_of_list.
% 1.70/1.89
% 1.70/1.89 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=4.
% 1.70/1.89
% 1.70/1.89 This ia a non-Horn set with equality. The strategy will be
% 1.70/1.89 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.70/1.89 deletion, with positive clauses in sos and nonpositive
% 1.70/1.89 clauses in usable.
% 1.70/1.89
% 1.70/1.89 dependent: set(knuth_bendix).
% 1.70/1.89 dependent: set(anl_eq).
% 1.70/1.89 dependent: set(para_from).
% 1.70/1.89 dependent: set(para_into).
% 1.70/1.89 dependent: clear(para_from_right).
% 1.70/1.89 dependent: clear(para_into_right).
% 1.70/1.89 dependent: set(para_from_vars).
% 1.70/1.89 dependent: set(eq_units_both_ways).
% 1.70/1.89 dependent: set(dynamic_demod_all).
% 1.70/1.89 dependent: set(dynamic_demod).
% 1.70/1.89 dependent: set(order_eq).
% 1.70/1.89 dependent: set(back_demod).
% 1.70/1.89 dependent: set(lrpo).
% 1.70/1.89 dependent: set(hyper_res).
% 1.70/1.89 dependent: set(unit_deletion).
% 1.70/1.89 dependent: set(factor).
% 1.70/1.89
% 1.70/1.89 ------------> process usable:
% 1.70/1.89 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.70/1.89 ** KEPT (pick-wt=14): 2 [] A!=unordered_pair(B,C)| -in(D,A)|D=B|D=C.
% 1.70/1.89 ** KEPT (pick-wt=11): 3 [] A!=unordered_pair(B,C)|in(D,A)|D!=B.
% 1.70/1.89 ** KEPT (pick-wt=11): 4 [] A!=unordered_pair(B,C)|in(D,A)|D!=C.
% 1.70/1.89 ** KEPT (pick-wt=17): 5 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=B.
% 1.70/1.89 ** KEPT (pick-wt=17): 6 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=C.
% 1.70/1.89 ** KEPT (pick-wt=9): 7 [] set_difference(singleton(A),B)!=empty_set|in(A,B).
% 1.70/1.89 ** KEPT (pick-wt=9): 8 [] set_difference(singleton(A),B)=empty_set| -in(A,B).
% 1.70/1.89 ** KEPT (pick-wt=2): 9 [] -empty($c2).
% 1.70/1.89 ** KEPT (pick-wt=8): 10 [] set_difference(singleton($c4),unordered_pair($c4,$c3))!=empty_set.
% 1.70/1.89
% 1.70/1.89 ------------> process sos:
% 1.70/1.89 ** KEPT (pick-wt=3): 13 [] A=A.
% 1.70/1.89 ** KEPT (pick-wt=7): 14 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.70/1.89 ** KEPT (pick-wt=23): 15 [] A=unordered_pair(B,C)|in($f1(B,C,A),A)|$f1(B,C,A)=B|$f1(B,C,A)=C.
% 1.70/1.89 ** KEPT (pick-wt=2): 16 [] empty(empty_set).
% 1.70/1.89 ** KEPT (pick-wt=2): 17 [] empty($c1).
% 1.70/1.89 Following clause subsumed by 13 during input processing: 0 [copy,13,flip.1] A=A.
% 1.70/1.89 Following clause subsumed by 14 during input processing: 0 [copy,14,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.70/1.89
% 1.70/1.89 ======= end of input processing =======
% 1.70/1.89
% 1.70/1.89 =========== start of search ===========
% 1.70/1.89
% 1.70/1.89 -------- PROOF --------
% 1.70/1.89
% 1.70/1.89 ----> UNIT CONFLICT at 0.00 sec ----> 40 [binary,38.1,10.1] $F.
% 1.70/1.89
% 1.70/1.89 Length of proof is 2. Level of proof is 2.
% 1.70/1.89
% 1.70/1.89 ---------------- PROOF ----------------
% 1.70/1.89 % SZS status Theorem
% 1.70/1.89 % SZS output start Refutation
% See solution above
% 1.70/1.89 ------------ end of proof -------------
% 1.70/1.89
% 1.70/1.89
% 1.70/1.89 Search stopped by max_proofs option.
% 1.70/1.89
% 1.70/1.89
% 1.70/1.89 Search stopped by max_proofs option.
% 1.70/1.89
% 1.70/1.89 ============ end of search ============
% 1.70/1.89
% 1.70/1.89 -------------- statistics -------------
% 1.70/1.89 clauses given 5
% 1.70/1.89 clauses generated 27
% 1.70/1.89 clauses kept 37
% 1.70/1.89 clauses forward subsumed 7
% 1.70/1.89 clauses back subsumed 0
% 1.70/1.89 Kbytes malloced 976
% 1.70/1.89
% 1.70/1.89 ----------- times (seconds) -----------
% 1.70/1.89 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.70/1.89 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.70/1.89 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.70/1.89
% 1.70/1.89 That finishes the proof of the theorem.
% 1.70/1.89
% 1.70/1.89 Process 14993 finished Wed Jul 27 10:35:37 2022
% 1.70/1.89 Otter interrupted
% 1.70/1.89 PROOF FOUND
%------------------------------------------------------------------------------