TSTP Solution File: SEU151+3 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU151+3 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n026.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 13:14:56 EDT 2022
% Result : Theorem 1.96s 2.14s
% Output : Refutation 1.96s
% Verified :
% SZS Type : Refutation
% Derivation depth : 2
% Number of leaves : 6
% Syntax : Number of clauses : 8 ( 6 unt; 1 nHn; 6 RR)
% Number of literals : 13 ( 9 equ; 6 neg)
% Maximal clause size : 4 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 4 con; 0-2 aty)
% Number of variables : 11 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( A != unordered_pair(B,C)
| ~ in(D,A)
| D = B
| D = C ),
file('SEU151+3.p',unknown),
[] ).
cnf(3,axiom,
( A != unordered_pair(B,C)
| in(D,A)
| D != B ),
file('SEU151+3.p',unknown),
[] ).
cnf(8,axiom,
dollar_c6 != dollar_c4,
file('SEU151+3.p',unknown),
[] ).
cnf(9,axiom,
dollar_c6 != dollar_c3,
file('SEU151+3.p',unknown),
[] ).
cnf(12,axiom,
A = A,
file('SEU151+3.p',unknown),
[] ).
cnf(16,axiom,
unordered_pair(dollar_c6,dollar_c5) = unordered_pair(dollar_c4,dollar_c3),
file('SEU151+3.p',unknown),
[] ).
cnf(20,plain,
in(A,unordered_pair(A,B)),
inference(hyper,[status(thm)],[12,3,12]),
[iquote('hyper,12,3,12')] ).
cnf(117,plain,
$false,
inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[16,2,20]),8,9]),
[iquote('hyper,16,2,20,unit_del,8,9')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11 % Problem : SEU151+3 : TPTP v8.1.0. Released v3.2.0.
% 0.11/0.11 % Command : otter-tptp-script %s
% 0.11/0.32 % Computer : n026.cluster.edu
% 0.11/0.32 % Model : x86_64 x86_64
% 0.11/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.32 % Memory : 8042.1875MB
% 0.11/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.32 % CPULimit : 300
% 0.11/0.32 % WCLimit : 300
% 0.11/0.32 % DateTime : Wed Jul 27 08:04:54 EDT 2022
% 0.11/0.33 % CPUTime :
% 1.96/2.14 ----- Otter 3.3f, August 2004 -----
% 1.96/2.14 The process was started by sandbox2 on n026.cluster.edu,
% 1.96/2.14 Wed Jul 27 08:04:54 2022
% 1.96/2.14 The command was "./otter". The process ID is 15418.
% 1.96/2.14
% 1.96/2.14 set(prolog_style_variables).
% 1.96/2.14 set(auto).
% 1.96/2.14 dependent: set(auto1).
% 1.96/2.14 dependent: set(process_input).
% 1.96/2.14 dependent: clear(print_kept).
% 1.96/2.14 dependent: clear(print_new_demod).
% 1.96/2.14 dependent: clear(print_back_demod).
% 1.96/2.14 dependent: clear(print_back_sub).
% 1.96/2.14 dependent: set(control_memory).
% 1.96/2.14 dependent: assign(max_mem, 12000).
% 1.96/2.14 dependent: assign(pick_given_ratio, 4).
% 1.96/2.14 dependent: assign(stats_level, 1).
% 1.96/2.14 dependent: assign(max_seconds, 10800).
% 1.96/2.14 clear(print_given).
% 1.96/2.14
% 1.96/2.14 formula_list(usable).
% 1.96/2.14 all A (A=A).
% 1.96/2.14 all A B (in(A,B)-> -in(B,A)).
% 1.96/2.14 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.96/2.14 all A B C (C=unordered_pair(A,B)<-> (all D (in(D,C)<->D=A|D=B))).
% 1.96/2.14 exists A empty(A).
% 1.96/2.14 exists A (-empty(A)).
% 1.96/2.14 -(all A B C D (-(unordered_pair(A,B)=unordered_pair(C,D)&A!=C&A!=D))).
% 1.96/2.14 end_of_list.
% 1.96/2.14
% 1.96/2.14 -------> usable clausifies to:
% 1.96/2.14
% 1.96/2.14 list(usable).
% 1.96/2.14 0 [] A=A.
% 1.96/2.14 0 [] -in(A,B)| -in(B,A).
% 1.96/2.14 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.96/2.14 0 [] C!=unordered_pair(A,B)| -in(D,C)|D=A|D=B.
% 1.96/2.14 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=A.
% 1.96/2.14 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=B.
% 1.96/2.14 0 [] C=unordered_pair(A,B)|in($f1(A,B,C),C)|$f1(A,B,C)=A|$f1(A,B,C)=B.
% 1.96/2.14 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=A.
% 1.96/2.14 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=B.
% 1.96/2.14 0 [] empty($c1).
% 1.96/2.14 0 [] -empty($c2).
% 1.96/2.14 0 [] unordered_pair($c6,$c5)=unordered_pair($c4,$c3).
% 1.96/2.14 0 [] $c6!=$c4.
% 1.96/2.14 0 [] $c6!=$c3.
% 1.96/2.14 end_of_list.
% 1.96/2.14
% 1.96/2.14 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=4.
% 1.96/2.14
% 1.96/2.14 This ia a non-Horn set with equality. The strategy will be
% 1.96/2.14 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.96/2.14 deletion, with positive clauses in sos and nonpositive
% 1.96/2.14 clauses in usable.
% 1.96/2.14
% 1.96/2.14 dependent: set(knuth_bendix).
% 1.96/2.14 dependent: set(anl_eq).
% 1.96/2.14 dependent: set(para_from).
% 1.96/2.14 dependent: set(para_into).
% 1.96/2.14 dependent: clear(para_from_right).
% 1.96/2.14 dependent: clear(para_into_right).
% 1.96/2.14 dependent: set(para_from_vars).
% 1.96/2.14 dependent: set(eq_units_both_ways).
% 1.96/2.14 dependent: set(dynamic_demod_all).
% 1.96/2.14 dependent: set(dynamic_demod).
% 1.96/2.14 dependent: set(order_eq).
% 1.96/2.14 dependent: set(back_demod).
% 1.96/2.14 dependent: set(lrpo).
% 1.96/2.14 dependent: set(hyper_res).
% 1.96/2.14 dependent: set(unit_deletion).
% 1.96/2.14 dependent: set(factor).
% 1.96/2.14
% 1.96/2.14 ------------> process usable:
% 1.96/2.14 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.96/2.14 ** KEPT (pick-wt=14): 2 [] A!=unordered_pair(B,C)| -in(D,A)|D=B|D=C.
% 1.96/2.14 ** KEPT (pick-wt=11): 3 [] A!=unordered_pair(B,C)|in(D,A)|D!=B.
% 1.96/2.14 ** KEPT (pick-wt=11): 4 [] A!=unordered_pair(B,C)|in(D,A)|D!=C.
% 1.96/2.14 ** KEPT (pick-wt=17): 5 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=B.
% 1.96/2.14 ** KEPT (pick-wt=17): 6 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=C.
% 1.96/2.14 ** KEPT (pick-wt=2): 7 [] -empty($c2).
% 1.96/2.14 ** KEPT (pick-wt=3): 8 [] $c6!=$c4.
% 1.96/2.14 ** KEPT (pick-wt=3): 9 [] $c6!=$c3.
% 1.96/2.14
% 1.96/2.14 ------------> process sos:
% 1.96/2.14 ** KEPT (pick-wt=3): 12 [] A=A.
% 1.96/2.14 ** KEPT (pick-wt=7): 13 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.96/2.14 ** KEPT (pick-wt=23): 14 [] A=unordered_pair(B,C)|in($f1(B,C,A),A)|$f1(B,C,A)=B|$f1(B,C,A)=C.
% 1.96/2.14 ** KEPT (pick-wt=2): 15 [] empty($c1).
% 1.96/2.14 ** KEPT (pick-wt=7): 16 [] unordered_pair($c6,$c5)=unordered_pair($c4,$c3).
% 1.96/2.14 ---> New Demodulator: 17 [new_demod,16] unordered_pair($c6,$c5)=unordered_pair($c4,$c3).
% 1.96/2.14 Following clause subsumed by 12 during input processing: 0 [copy,12,flip.1] A=A.
% 1.96/2.14 Following clause subsumed by 13 during input processing: 0 [copy,13,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.96/2.14 >>>> Starting back demodulation with 17.
% 1.96/2.14
% 1.96/2.14 ======= end of input processing =======
% 1.96/2.14
% 1.96/2.14 =========== start of search ===========
% 1.96/2.14
% 1.96/2.14 -------- PROOF --------
% 1.96/2.14
% 1.96/2.14 -----> EMPTY CLAUSE at 0.01 sec ----> 117 [hyper,16,2,20,unit_del,8,9] $F.
% 1.96/2.14
% 1.96/2.14 Length of proof is 1. Level of proof is 1.
% 1.96/2.14
% 1.96/2.14 ---------------- PROOF ----------------
% 1.96/2.14 % SZS status Theorem
% 1.96/2.14 % SZS output start Refutation
% See solution above
% 1.96/2.15 ------------ end of proof -------------
% 1.96/2.15
% 1.96/2.15
% 1.96/2.15 Search stopped by max_proofs option.
% 1.96/2.15
% 1.96/2.15
% 1.96/2.15 Search stopped by max_proofs option.
% 1.96/2.15
% 1.96/2.15 ============ end of search ============
% 1.96/2.15
% 1.96/2.15 -------------- statistics -------------
% 1.96/2.15 clauses given 7
% 1.96/2.15 clauses generated 191
% 1.96/2.15 clauses kept 115
% 1.96/2.15 clauses forward subsumed 85
% 1.96/2.15 clauses back subsumed 0
% 1.96/2.15 Kbytes malloced 976
% 1.96/2.15
% 1.96/2.15 ----------- times (seconds) -----------
% 1.96/2.15 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.96/2.15 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.96/2.15 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.96/2.15
% 1.96/2.15 That finishes the proof of the theorem.
% 1.96/2.15
% 1.96/2.15 Process 15418 finished Wed Jul 27 08:04:56 2022
% 1.96/2.15 Otter interrupted
% 1.96/2.15 PROOF FOUND
%------------------------------------------------------------------------------