TSTP Solution File: SEU167+3 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU167+3 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n021.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 13:15:01 EDT 2022
% Result : Theorem 1.62s 1.80s
% Output : Refutation 1.62s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 6
% Syntax : Number of clauses : 10 ( 7 unt; 0 nHn; 6 RR)
% Number of literals : 14 ( 0 equ; 5 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 2 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 4 con; 0-2 aty)
% Number of variables : 11 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( ~ subset(A,B)
| subset(cartesian_product2(A,C),cartesian_product2(B,C)) ),
file('SEU167+3.p',unknown),
[] ).
cnf(3,axiom,
( ~ subset(A,B)
| subset(cartesian_product2(C,A),cartesian_product2(C,B)) ),
file('SEU167+3.p',unknown),
[] ).
cnf(4,axiom,
~ subset(cartesian_product2(dollar_c6,dollar_c4),cartesian_product2(dollar_c5,dollar_c3)),
file('SEU167+3.p',unknown),
[] ).
cnf(5,axiom,
( ~ subset(A,B)
| ~ subset(B,C)
| subset(A,C) ),
file('SEU167+3.p',unknown),
[] ).
cnf(8,axiom,
subset(dollar_c6,dollar_c5),
file('SEU167+3.p',unknown),
[] ).
cnf(9,axiom,
subset(dollar_c4,dollar_c3),
file('SEU167+3.p',unknown),
[] ).
cnf(11,plain,
subset(cartesian_product2(dollar_c6,A),cartesian_product2(dollar_c5,A)),
inference(hyper,[status(thm)],[8,2]),
[iquote('hyper,8,2')] ).
cnf(12,plain,
subset(cartesian_product2(A,dollar_c4),cartesian_product2(A,dollar_c3)),
inference(hyper,[status(thm)],[9,3]),
[iquote('hyper,9,3')] ).
cnf(19,plain,
subset(cartesian_product2(dollar_c6,dollar_c4),cartesian_product2(dollar_c5,dollar_c3)),
inference(hyper,[status(thm)],[12,5,11]),
[iquote('hyper,12,5,11')] ).
cnf(20,plain,
$false,
inference(binary,[status(thm)],[19,4]),
[iquote('binary,19.1,4.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SEU167+3 : TPTP v8.1.0. Released v3.2.0.
% 0.11/0.12 % Command : otter-tptp-script %s
% 0.13/0.33 % Computer : n021.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Wed Jul 27 07:35:08 EDT 2022
% 0.13/0.33 % CPUTime :
% 1.62/1.80 ----- Otter 3.3f, August 2004 -----
% 1.62/1.80 The process was started by sandbox on n021.cluster.edu,
% 1.62/1.80 Wed Jul 27 07:35:08 2022
% 1.62/1.80 The command was "./otter". The process ID is 20402.
% 1.62/1.80
% 1.62/1.80 set(prolog_style_variables).
% 1.62/1.80 set(auto).
% 1.62/1.80 dependent: set(auto1).
% 1.62/1.80 dependent: set(process_input).
% 1.62/1.80 dependent: clear(print_kept).
% 1.62/1.80 dependent: clear(print_new_demod).
% 1.62/1.80 dependent: clear(print_back_demod).
% 1.62/1.80 dependent: clear(print_back_sub).
% 1.62/1.80 dependent: set(control_memory).
% 1.62/1.80 dependent: assign(max_mem, 12000).
% 1.62/1.80 dependent: assign(pick_given_ratio, 4).
% 1.62/1.80 dependent: assign(stats_level, 1).
% 1.62/1.80 dependent: assign(max_seconds, 10800).
% 1.62/1.80 clear(print_given).
% 1.62/1.80
% 1.62/1.80 formula_list(usable).
% 1.62/1.80 exists A empty(A).
% 1.62/1.80 exists A (-empty(A)).
% 1.62/1.80 all A B subset(A,A).
% 1.62/1.80 all A B C (subset(A,B)->subset(cartesian_product2(A,C),cartesian_product2(B,C))&subset(cartesian_product2(C,A),cartesian_product2(C,B))).
% 1.62/1.80 -(all A B C D (subset(A,B)&subset(C,D)->subset(cartesian_product2(A,C),cartesian_product2(B,D)))).
% 1.62/1.80 all A B C (subset(A,B)&subset(B,C)->subset(A,C)).
% 1.62/1.80 end_of_list.
% 1.62/1.80
% 1.62/1.80 -------> usable clausifies to:
% 1.62/1.80
% 1.62/1.80 list(usable).
% 1.62/1.80 0 [] empty($c1).
% 1.62/1.80 0 [] -empty($c2).
% 1.62/1.80 0 [] subset(A,A).
% 1.62/1.80 0 [] -subset(A,B)|subset(cartesian_product2(A,C),cartesian_product2(B,C)).
% 1.62/1.80 0 [] -subset(A,B)|subset(cartesian_product2(C,A),cartesian_product2(C,B)).
% 1.62/1.80 0 [] subset($c6,$c5).
% 1.62/1.80 0 [] subset($c4,$c3).
% 1.62/1.80 0 [] -subset(cartesian_product2($c6,$c4),cartesian_product2($c5,$c3)).
% 1.62/1.80 0 [] -subset(A,B)| -subset(B,C)|subset(A,C).
% 1.62/1.80 end_of_list.
% 1.62/1.80
% 1.62/1.80 SCAN INPUT: prop=0, horn=1, equality=0, symmetry=0, max_lits=3.
% 1.62/1.80
% 1.62/1.80 This is a Horn set without equality. The strategy will
% 1.62/1.80 be hyperresolution, with satellites in sos and nuclei
% 1.62/1.80 in usable.
% 1.62/1.80
% 1.62/1.80 dependent: set(hyper_res).
% 1.62/1.80 dependent: clear(order_hyper).
% 1.62/1.80
% 1.62/1.80 ------------> process usable:
% 1.62/1.80 ** KEPT (pick-wt=2): 1 [] -empty($c2).
% 1.62/1.80 ** KEPT (pick-wt=10): 2 [] -subset(A,B)|subset(cartesian_product2(A,C),cartesian_product2(B,C)).
% 1.62/1.80 ** KEPT (pick-wt=10): 3 [] -subset(A,B)|subset(cartesian_product2(C,A),cartesian_product2(C,B)).
% 1.62/1.80 ** KEPT (pick-wt=7): 4 [] -subset(cartesian_product2($c6,$c4),cartesian_product2($c5,$c3)).
% 1.62/1.80 ** KEPT (pick-wt=9): 5 [] -subset(A,B)| -subset(B,C)|subset(A,C).
% 1.62/1.80
% 1.62/1.80 ------------> process sos:
% 1.62/1.80 ** KEPT (pick-wt=2): 6 [] empty($c1).
% 1.62/1.80 ** KEPT (pick-wt=3): 7 [] subset(A,A).
% 1.62/1.80 ** KEPT (pick-wt=3): 8 [] subset($c6,$c5).
% 1.62/1.80 ** KEPT (pick-wt=3): 9 [] subset($c4,$c3).
% 1.62/1.80
% 1.62/1.80 ======= end of input processing =======
% 1.62/1.80
% 1.62/1.80 =========== start of search ===========
% 1.62/1.80
% 1.62/1.80 -------- PROOF --------
% 1.62/1.80
% 1.62/1.80 ----> UNIT CONFLICT at 0.00 sec ----> 20 [binary,19.1,4.1] $F.
% 1.62/1.80
% 1.62/1.80 Length of proof is 3. Level of proof is 2.
% 1.62/1.80
% 1.62/1.80 ---------------- PROOF ----------------
% 1.62/1.80 % SZS status Theorem
% 1.62/1.80 % SZS output start Refutation
% See solution above
% 1.62/1.80 ------------ end of proof -------------
% 1.62/1.80
% 1.62/1.80
% 1.62/1.80 Search stopped by max_proofs option.
% 1.62/1.80
% 1.62/1.80
% 1.62/1.80 Search stopped by max_proofs option.
% 1.62/1.80
% 1.62/1.80 ============ end of search ============
% 1.62/1.80
% 1.62/1.80 -------------- statistics -------------
% 1.62/1.80 clauses given 7
% 1.62/1.80 clauses generated 23
% 1.62/1.80 clauses kept 19
% 1.62/1.80 clauses forward subsumed 13
% 1.62/1.80 clauses back subsumed 0
% 1.62/1.80 Kbytes malloced 976
% 1.62/1.80
% 1.62/1.80 ----------- times (seconds) -----------
% 1.62/1.80 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.62/1.80 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.62/1.80 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.62/1.80
% 1.62/1.80 That finishes the proof of the theorem.
% 1.62/1.80
% 1.62/1.80 Process 20402 finished Wed Jul 27 07:35:09 2022
% 1.62/1.80 Otter interrupted
% 1.62/1.80 PROOF FOUND
%------------------------------------------------------------------------------