TSTP Solution File: SWW417-1 by Otter---3.3
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%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SWW417-1 : TPTP v8.1.0. Released v5.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n014.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 13:22:31 EDT 2022
% Result : Unsatisfiable 1.72s 1.92s
% Output : Refutation 1.72s
% Verified :
% SZS Type : Refutation
% Derivation depth : 2
% Number of leaves : 5
% Syntax : Number of clauses : 7 ( 6 unt; 1 nHn; 6 RR)
% Number of literals : 9 ( 3 equ; 3 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 4 ( 2 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 7 ( 7 usr; 4 con; 0-2 aty)
% Number of variables : 7 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(6,axiom,
( ~ heap(sep(next(A,B),sep(lseg(B,C),D)))
| A = B
| heap(sep(lseg(A,C),D)) ),
file('SWW417-1.p',unknown),
[] ).
cnf(12,axiom,
x2 != x1,
file('SWW417-1.p',unknown),
[] ).
cnf(13,axiom,
~ heap(sep(lseg(x1,nil),emp)),
file('SWW417-1.p',unknown),
[] ).
cnf(16,axiom,
sep(A,sep(B,C)) = sep(B,sep(A,C)),
file('SWW417-1.p',unknown),
[] ).
cnf(19,axiom,
heap(sep(lseg(x2,nil),sep(next(x1,x2),emp))),
file('SWW417-1.p',unknown),
[] ).
cnf(163,plain,
heap(sep(next(x1,x2),sep(lseg(x2,nil),emp))),
inference(para_from,[status(thm),theory(equality)],[16,19]),
[iquote('para_from,16.1.1,19.1.1')] ).
cnf(218,plain,
$false,
inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[163,6]),12,13]),
[iquote('hyper,163,6,unit_del,12,13')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : SWW417-1 : TPTP v8.1.0. Released v5.2.0.
% 0.06/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n014.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 02:51:54 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.72/1.92 ----- Otter 3.3f, August 2004 -----
% 1.72/1.92 The process was started by sandbox on n014.cluster.edu,
% 1.72/1.92 Wed Jul 27 02:51:54 2022
% 1.72/1.92 The command was "./otter". The process ID is 18820.
% 1.72/1.92
% 1.72/1.92 set(prolog_style_variables).
% 1.72/1.92 set(auto).
% 1.72/1.92 dependent: set(auto1).
% 1.72/1.92 dependent: set(process_input).
% 1.72/1.92 dependent: clear(print_kept).
% 1.72/1.92 dependent: clear(print_new_demod).
% 1.72/1.92 dependent: clear(print_back_demod).
% 1.72/1.92 dependent: clear(print_back_sub).
% 1.72/1.92 dependent: set(control_memory).
% 1.72/1.92 dependent: assign(max_mem, 12000).
% 1.72/1.92 dependent: assign(pick_given_ratio, 4).
% 1.72/1.92 dependent: assign(stats_level, 1).
% 1.72/1.92 dependent: assign(max_seconds, 10800).
% 1.72/1.92 clear(print_given).
% 1.72/1.92
% 1.72/1.92 list(usable).
% 1.72/1.92 0 [] A=A.
% 1.72/1.92 0 [] sep(S,sep(T,Sigma))=sep(T,sep(S,Sigma)).
% 1.72/1.92 0 [] sep(lseg(X,X),Sigma)=Sigma.
% 1.72/1.92 0 [] -heap(sep(next(nil,Y),Sigma)).
% 1.72/1.92 0 [] -heap(sep(lseg(nil,Y),Sigma))|Y=nil.
% 1.72/1.92 0 [] -heap(sep(next(X,Y),sep(next(X,Z),Sigma))).
% 1.72/1.92 0 [] -heap(sep(next(X,Y),sep(lseg(X,Z),Sigma)))|X=Z.
% 1.72/1.92 0 [] -heap(sep(lseg(X,Y),sep(lseg(X,Z),Sigma)))|X=Y|X=Z.
% 1.72/1.92 0 [] -heap(sep(next(X,Y),sep(lseg(Y,Z),Sigma)))|X=Y|heap(sep(lseg(X,Z),Sigma)).
% 1.72/1.92 0 [] -heap(sep(lseg(X,Y),sep(lseg(Y,nil),Sigma)))|heap(sep(lseg(X,nil),Sigma)).
% 1.72/1.92 0 [] -heap(sep(lseg(X,Y),sep(lseg(Y,Z),sep(next(Z,W),Sigma))))|heap(sep(lseg(X,Z),sep(next(Z,W),Sigma))).
% 1.72/1.92 0 [] -heap(sep(lseg(X,Y),sep(lseg(Y,Z),sep(lseg(Z,W),Sigma))))|Z=W|heap(sep(lseg(X,Z),sep(lseg(Z,W),Sigma))).
% 1.72/1.92 0 [] nil!=x1.
% 1.72/1.92 0 [] x2!=x1.
% 1.72/1.92 0 [] heap(sep(lseg(x2,nil),sep(next(x1,x2),emp))).
% 1.72/1.92 0 [] -heap(sep(lseg(x1,nil),emp)).
% 1.72/1.92 end_of_list.
% 1.72/1.92
% 1.72/1.92 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.72/1.92
% 1.72/1.92 This ia a non-Horn set with equality. The strategy will be
% 1.72/1.92 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.72/1.92 deletion, with positive clauses in sos and nonpositive
% 1.72/1.92 clauses in usable.
% 1.72/1.92
% 1.72/1.92 dependent: set(knuth_bendix).
% 1.72/1.92 dependent: set(anl_eq).
% 1.72/1.92 dependent: set(para_from).
% 1.72/1.92 dependent: set(para_into).
% 1.72/1.92 dependent: clear(para_from_right).
% 1.72/1.92 dependent: clear(para_into_right).
% 1.72/1.92 dependent: set(para_from_vars).
% 1.72/1.92 dependent: set(eq_units_both_ways).
% 1.72/1.92 dependent: set(dynamic_demod_all).
% 1.72/1.92 dependent: set(dynamic_demod).
% 1.72/1.92 dependent: set(order_eq).
% 1.72/1.92 dependent: set(back_demod).
% 1.72/1.92 dependent: set(lrpo).
% 1.72/1.92 dependent: set(hyper_res).
% 1.72/1.92 dependent: set(unit_deletion).
% 1.72/1.92 dependent: set(factor).
% 1.72/1.92
% 1.72/1.92 ------------> process usable:
% 1.72/1.92 ** KEPT (pick-wt=6): 1 [] -heap(sep(next(nil,A),B)).
% 1.72/1.92 ** KEPT (pick-wt=9): 2 [] -heap(sep(lseg(nil,A),B))|A=nil.
% 1.72/1.92 ** KEPT (pick-wt=10): 3 [] -heap(sep(next(A,B),sep(next(A,C),D))).
% 1.72/1.92 ** KEPT (pick-wt=13): 4 [] -heap(sep(next(A,B),sep(lseg(A,C),D)))|A=C.
% 1.72/1.92 ** KEPT (pick-wt=16): 5 [] -heap(sep(lseg(A,B),sep(lseg(A,C),D)))|A=B|A=C.
% 1.72/1.92 ** KEPT (pick-wt=19): 6 [] -heap(sep(next(A,B),sep(lseg(B,C),D)))|A=B|heap(sep(lseg(A,C),D)).
% 1.72/1.92 ** KEPT (pick-wt=16): 7 [] -heap(sep(lseg(A,B),sep(lseg(B,nil),C)))|heap(sep(lseg(A,nil),C)).
% 1.72/1.92 ** KEPT (pick-wt=24): 8 [] -heap(sep(lseg(A,B),sep(lseg(B,C),sep(next(C,D),E))))|heap(sep(lseg(A,C),sep(next(C,D),E))).
% 1.72/1.92 ** KEPT (pick-wt=27): 9 [] -heap(sep(lseg(A,B),sep(lseg(B,C),sep(lseg(C,D),E))))|C=D|heap(sep(lseg(A,C),sep(lseg(C,D),E))).
% 1.72/1.92 ** KEPT (pick-wt=3): 11 [copy,10,flip.1] x1!=nil.
% 1.72/1.92 ** KEPT (pick-wt=3): 12 [] x2!=x1.
% 1.72/1.92 ** KEPT (pick-wt=6): 13 [] -heap(sep(lseg(x1,nil),emp)).
% 1.72/1.92
% 1.72/1.92 ------------> process sos:
% 1.72/1.92 ** KEPT (pick-wt=3): 15 [] A=A.
% 1.72/1.92 ** KEPT (pick-wt=11): 16 [] sep(A,sep(B,C))=sep(B,sep(A,C)).
% 1.72/1.92 ** KEPT (pick-wt=7): 17 [] sep(lseg(A,A),B)=B.
% 1.72/1.92 ---> New Demodulator: 18 [new_demod,17] sep(lseg(A,A),B)=B.
% 1.72/1.92 ** KEPT (pick-wt=10): 19 [] heap(sep(lseg(x2,nil),sep(next(x1,x2),emp))).
% 1.72/1.92 Following clause subsumed by 15 during input processing: 0 [copy,15,flip.1] A=A.
% 1.72/1.92 Following clause subsumed by 16 during input processing: 0 [copy,16,flip.1] sep(A,sep(B,C))=sep(B,sep(A,C)).
% 1.72/1.92 >>>> Starting back demodulation with 18.
% 1.72/1.92
% 1.72/1.92 ======= end of input processing =======
% 1.72/1.92
% 1.72/1.92 =========== start of search ===========
% 1.72/1.92
% 1.72/1.92 -------- PROOF --------
% 1.72/1.92
% 1.72/1.92 -----> EMPTY CLAUSE at 0.01 sec ----> 218 [hyper,163,6,unit_del,12,13] $F.
% 1.72/1.92
% 1.72/1.92 Length of proof is 1. Level of proof is 1.
% 1.72/1.92
% 1.72/1.92 ---------------- PROOF ----------------
% 1.72/1.92 % SZS status Unsatisfiable
% 1.72/1.92 % SZS output start Refutation
% See solution above
% 1.72/1.92 ------------ end of proof -------------
% 1.72/1.92
% 1.72/1.92
% 1.72/1.92 Search stopped by max_proofs option.
% 1.72/1.92
% 1.72/1.92
% 1.72/1.92 Search stopped by max_proofs option.
% 1.72/1.92
% 1.72/1.92 ============ end of search ============
% 1.72/1.92
% 1.72/1.92 -------------- statistics -------------
% 1.72/1.92 clauses given 8
% 1.72/1.92 clauses generated 277
% 1.72/1.92 clauses kept 215
% 1.72/1.92 clauses forward subsumed 71
% 1.72/1.92 clauses back subsumed 0
% 1.72/1.92 Kbytes malloced 2929
% 1.72/1.92
% 1.72/1.92 ----------- times (seconds) -----------
% 1.72/1.92 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.72/1.92 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.72/1.92 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.72/1.92
% 1.72/1.92 That finishes the proof of the theorem.
% 1.72/1.92
% 1.72/1.92 Process 18820 finished Wed Jul 27 02:51:56 2022
% 1.72/1.92 Otter interrupted
% 1.72/1.92 PROOF FOUND
%------------------------------------------------------------------------------