TSTP Solution File: SET912+1 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET912+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n022.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 13:14:31 EDT 2022
% Result : Theorem 1.73s 1.95s
% Output : Refutation 1.73s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 5
% Syntax : Number of clauses : 8 ( 6 unt; 0 nHn; 8 RR)
% Number of literals : 11 ( 3 equ; 4 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 : 5 ( 5 usr; 3 con; 0-2 aty)
% Number of variables : 5 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(3,axiom,
( ~ subset(A,B)
| set_intersection2(A,B) = A ),
file('SET912+1.p',unknown),
[] ).
cnf(6,axiom,
( subset(unordered_pair(A,B),C)
| ~ in(A,C)
| ~ in(B,C) ),
file('SET912+1.p',unknown),
[] ).
cnf(7,axiom,
set_intersection2(unordered_pair(dollar_c5,dollar_c3),dollar_c4) != unordered_pair(dollar_c5,dollar_c3),
file('SET912+1.p',unknown),
[] ).
cnf(15,axiom,
in(dollar_c5,dollar_c4),
file('SET912+1.p',unknown),
[] ).
cnf(16,axiom,
in(dollar_c3,dollar_c4),
file('SET912+1.p',unknown),
[] ).
cnf(21,plain,
subset(unordered_pair(dollar_c5,dollar_c3),dollar_c4),
inference(hyper,[status(thm)],[16,6,15]),
[iquote('hyper,16,6,15')] ).
cnf(33,plain,
set_intersection2(unordered_pair(dollar_c5,dollar_c3),dollar_c4) = unordered_pair(dollar_c5,dollar_c3),
inference(hyper,[status(thm)],[21,3]),
[iquote('hyper,21,3')] ).
cnf(35,plain,
$false,
inference(binary,[status(thm)],[33,7]),
[iquote('binary,33.1,7.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : SET912+1 : TPTP v8.1.0. Released v3.2.0.
% 0.12/0.13 % Command : otter-tptp-script %s
% 0.13/0.34 % Computer : n022.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 10:36:05 EDT 2022
% 0.13/0.34 % CPUTime :
% 1.73/1.95 ----- Otter 3.3f, August 2004 -----
% 1.73/1.95 The process was started by sandbox2 on n022.cluster.edu,
% 1.73/1.95 Wed Jul 27 10:36:05 2022
% 1.73/1.95 The command was "./otter". The process ID is 16060.
% 1.73/1.95
% 1.73/1.95 set(prolog_style_variables).
% 1.73/1.95 set(auto).
% 1.73/1.95 dependent: set(auto1).
% 1.73/1.95 dependent: set(process_input).
% 1.73/1.95 dependent: clear(print_kept).
% 1.73/1.95 dependent: clear(print_new_demod).
% 1.73/1.95 dependent: clear(print_back_demod).
% 1.73/1.95 dependent: clear(print_back_sub).
% 1.73/1.95 dependent: set(control_memory).
% 1.73/1.95 dependent: assign(max_mem, 12000).
% 1.73/1.95 dependent: assign(pick_given_ratio, 4).
% 1.73/1.95 dependent: assign(stats_level, 1).
% 1.73/1.95 dependent: assign(max_seconds, 10800).
% 1.73/1.95 clear(print_given).
% 1.73/1.95
% 1.73/1.95 formula_list(usable).
% 1.73/1.95 all A (A=A).
% 1.73/1.95 all A B (in(A,B)-> -in(B,A)).
% 1.73/1.95 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.73/1.95 all A B (set_intersection2(A,B)=set_intersection2(B,A)).
% 1.73/1.95 all A B (set_intersection2(A,A)=A).
% 1.73/1.95 exists A empty(A).
% 1.73/1.95 exists A (-empty(A)).
% 1.73/1.95 all A B subset(A,A).
% 1.73/1.95 all A B (subset(A,B)->set_intersection2(A,B)=A).
% 1.73/1.95 all A B C (subset(unordered_pair(A,B),C)<->in(A,C)&in(B,C)).
% 1.73/1.95 -(all A B C (in(A,B)&in(C,B)->set_intersection2(unordered_pair(A,C),B)=unordered_pair(A,C))).
% 1.73/1.95 end_of_list.
% 1.73/1.95
% 1.73/1.95 -------> usable clausifies to:
% 1.73/1.95
% 1.73/1.95 list(usable).
% 1.73/1.95 0 [] A=A.
% 1.73/1.95 0 [] -in(A,B)| -in(B,A).
% 1.73/1.95 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.73/1.95 0 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.73/1.95 0 [] set_intersection2(A,A)=A.
% 1.73/1.95 0 [] empty($c1).
% 1.73/1.95 0 [] -empty($c2).
% 1.73/1.95 0 [] subset(A,A).
% 1.73/1.95 0 [] -subset(A,B)|set_intersection2(A,B)=A.
% 1.73/1.95 0 [] -subset(unordered_pair(A,B),C)|in(A,C).
% 1.73/1.95 0 [] -subset(unordered_pair(A,B),C)|in(B,C).
% 1.73/1.95 0 [] subset(unordered_pair(A,B),C)| -in(A,C)| -in(B,C).
% 1.73/1.95 0 [] in($c5,$c4).
% 1.73/1.95 0 [] in($c3,$c4).
% 1.73/1.95 0 [] set_intersection2(unordered_pair($c5,$c3),$c4)!=unordered_pair($c5,$c3).
% 1.73/1.95 end_of_list.
% 1.73/1.95
% 1.73/1.95 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=3.
% 1.73/1.95
% 1.73/1.95 This is a Horn set with equality. The strategy will be
% 1.73/1.95 Knuth-Bendix and hyper_res, with positive clauses in
% 1.73/1.95 sos and nonpositive clauses in usable.
% 1.73/1.95
% 1.73/1.95 dependent: set(knuth_bendix).
% 1.73/1.95 dependent: set(anl_eq).
% 1.73/1.95 dependent: set(para_from).
% 1.73/1.95 dependent: set(para_into).
% 1.73/1.95 dependent: clear(para_from_right).
% 1.73/1.95 dependent: clear(para_into_right).
% 1.73/1.95 dependent: set(para_from_vars).
% 1.73/1.95 dependent: set(eq_units_both_ways).
% 1.73/1.95 dependent: set(dynamic_demod_all).
% 1.73/1.95 dependent: set(dynamic_demod).
% 1.73/1.95 dependent: set(order_eq).
% 1.73/1.95 dependent: set(back_demod).
% 1.73/1.95 dependent: set(lrpo).
% 1.73/1.95 dependent: set(hyper_res).
% 1.73/1.95 dependent: clear(order_hyper).
% 1.73/1.95
% 1.73/1.95 ------------> process usable:
% 1.73/1.95 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.73/1.95 ** KEPT (pick-wt=2): 2 [] -empty($c2).
% 1.73/1.95 ** KEPT (pick-wt=8): 3 [] -subset(A,B)|set_intersection2(A,B)=A.
% 1.73/1.95 ** KEPT (pick-wt=8): 4 [] -subset(unordered_pair(A,B),C)|in(A,C).
% 1.73/1.95 ** KEPT (pick-wt=8): 5 [] -subset(unordered_pair(A,B),C)|in(B,C).
% 1.73/1.95 ** KEPT (pick-wt=11): 6 [] subset(unordered_pair(A,B),C)| -in(A,C)| -in(B,C).
% 1.73/1.95 ** KEPT (pick-wt=9): 7 [] set_intersection2(unordered_pair($c5,$c3),$c4)!=unordered_pair($c5,$c3).
% 1.73/1.95
% 1.73/1.95 ------------> process sos:
% 1.73/1.95 ** KEPT (pick-wt=3): 8 [] A=A.
% 1.73/1.95 ** KEPT (pick-wt=7): 9 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.73/1.95 ** KEPT (pick-wt=7): 10 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.73/1.95 ** KEPT (pick-wt=5): 11 [] set_intersection2(A,A)=A.
% 1.73/1.95 ---> New Demodulator: 12 [new_demod,11] set_intersection2(A,A)=A.
% 1.73/1.95 ** KEPT (pick-wt=2): 13 [] empty($c1).
% 1.73/1.95 ** KEPT (pick-wt=3): 14 [] subset(A,A).
% 1.73/1.95 ** KEPT (pick-wt=3): 15 [] in($c5,$c4).
% 1.73/1.95 ** KEPT (pick-wt=3): 16 [] in($c3,$c4).
% 1.73/1.95 Following clause subsumed by 8 during input processing: 0 [copy,8,flip.1] A=A.
% 1.73/1.95 Following clause subsumed by 9 during input processing: 0 [copy,9,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.73/1.95 Following clause subsumed by 10 during input processing: 0 [copy,10,flip.1] set_intersection2(A,B)=set_intersection2(B,A).
% 1.73/1.95 >>>> Starting back demodulation with 12.
% 1.73/1.95
% 1.73/1.95 ======= end of input processing =======
% 1.73/1.95
% 1.73/1.95 =========== start of search ===========
% 1.73/1.95
% 1.73/1.95 -------- PROOF --------
% 1.73/1.95
% 1.73/1.95 ----> UNIT CONFLICT at 0.00 sec ----> 35 [binary,33.1,7.1] $F.
% 1.73/1.95
% 1.73/1.95 Length of proof is 2. Level of proof is 2.
% 1.73/1.95
% 1.73/1.95 ---------------- PROOF ----------------
% 1.73/1.95 % SZS status Theorem
% 1.73/1.95 % SZS output start Refutation
% See solution above
% 1.73/1.95 ------------ end of proof -------------
% 1.73/1.95
% 1.73/1.95
% 1.73/1.95 Search stopped by max_proofs option.
% 1.73/1.95
% 1.73/1.95
% 1.73/1.95 Search stopped by max_proofs option.
% 1.73/1.95
% 1.73/1.95 ============ end of search ============
% 1.73/1.95
% 1.73/1.95 -------------- statistics -------------
% 1.73/1.95 clauses given 13
% 1.73/1.95 clauses generated 45
% 1.73/1.95 clauses kept 30
% 1.73/1.95 clauses forward subsumed 33
% 1.73/1.95 clauses back subsumed 0
% 1.73/1.95 Kbytes malloced 976
% 1.73/1.95
% 1.73/1.95 ----------- times (seconds) -----------
% 1.73/1.95 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.73/1.95 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.73/1.95 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.73/1.95
% 1.73/1.95 That finishes the proof of the theorem.
% 1.73/1.95
% 1.73/1.95 Process 16060 finished Wed Jul 27 10:36:07 2022
% 1.73/1.95 Otter interrupted
% 1.73/1.95 PROOF FOUND
%------------------------------------------------------------------------------