TSTP Solution File: SEU151+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU151+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n017.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 1.66s 1.88s
% Output : Refutation 1.66s
% 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+1.p',unknown),
[] ).
cnf(3,axiom,
( A != unordered_pair(B,C)
| in(D,A)
| D != B ),
file('SEU151+1.p',unknown),
[] ).
cnf(7,axiom,
dollar_c4 != dollar_c2,
file('SEU151+1.p',unknown),
[] ).
cnf(8,axiom,
dollar_c4 != dollar_c1,
file('SEU151+1.p',unknown),
[] ).
cnf(11,axiom,
A = A,
file('SEU151+1.p',unknown),
[] ).
cnf(14,axiom,
unordered_pair(dollar_c4,dollar_c3) = unordered_pair(dollar_c2,dollar_c1),
file('SEU151+1.p',unknown),
[] ).
cnf(18,plain,
in(A,unordered_pair(A,B)),
inference(hyper,[status(thm)],[11,3,11]),
[iquote('hyper,11,3,11')] ).
cnf(36,plain,
$false,
inference(unit_del,[status(thm)],[inference(hyper,[status(thm)],[14,2,18]),7,8]),
[iquote('hyper,14,2,18,unit_del,7,8')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.11 % Problem : SEU151+1 : TPTP v8.1.0. Released v3.3.0.
% 0.04/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n017.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 06:56:18 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.66/1.88 ----- Otter 3.3f, August 2004 -----
% 1.66/1.88 The process was started by sandbox on n017.cluster.edu,
% 1.66/1.88 Wed Jul 27 06:56:18 2022
% 1.66/1.88 The command was "./otter". The process ID is 24775.
% 1.66/1.88
% 1.66/1.88 set(prolog_style_variables).
% 1.66/1.88 set(auto).
% 1.66/1.88 dependent: set(auto1).
% 1.66/1.88 dependent: set(process_input).
% 1.66/1.88 dependent: clear(print_kept).
% 1.66/1.88 dependent: clear(print_new_demod).
% 1.66/1.88 dependent: clear(print_back_demod).
% 1.66/1.88 dependent: clear(print_back_sub).
% 1.66/1.88 dependent: set(control_memory).
% 1.66/1.88 dependent: assign(max_mem, 12000).
% 1.66/1.88 dependent: assign(pick_given_ratio, 4).
% 1.66/1.88 dependent: assign(stats_level, 1).
% 1.66/1.88 dependent: assign(max_seconds, 10800).
% 1.66/1.88 clear(print_given).
% 1.66/1.88
% 1.66/1.88 formula_list(usable).
% 1.66/1.88 all A (A=A).
% 1.66/1.88 all A B (in(A,B)-> -in(B,A)).
% 1.66/1.88 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.66/1.88 all A B C (C=unordered_pair(A,B)<-> (all D (in(D,C)<->D=A|D=B))).
% 1.66/1.88 $T.
% 1.66/1.88 -(all A B C D (-(unordered_pair(A,B)=unordered_pair(C,D)&A!=C&A!=D))).
% 1.66/1.88 end_of_list.
% 1.66/1.88
% 1.66/1.88 -------> usable clausifies to:
% 1.66/1.88
% 1.66/1.88 list(usable).
% 1.66/1.88 0 [] A=A.
% 1.66/1.88 0 [] -in(A,B)| -in(B,A).
% 1.66/1.88 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.66/1.88 0 [] C!=unordered_pair(A,B)| -in(D,C)|D=A|D=B.
% 1.66/1.88 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=A.
% 1.66/1.88 0 [] C!=unordered_pair(A,B)|in(D,C)|D!=B.
% 1.66/1.88 0 [] C=unordered_pair(A,B)|in($f1(A,B,C),C)|$f1(A,B,C)=A|$f1(A,B,C)=B.
% 1.66/1.88 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=A.
% 1.66/1.88 0 [] C=unordered_pair(A,B)| -in($f1(A,B,C),C)|$f1(A,B,C)!=B.
% 1.66/1.88 0 [] $T.
% 1.66/1.88 0 [] unordered_pair($c4,$c3)=unordered_pair($c2,$c1).
% 1.66/1.88 0 [] $c4!=$c2.
% 1.66/1.88 0 [] $c4!=$c1.
% 1.66/1.88 end_of_list.
% 1.66/1.88
% 1.66/1.88 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=4.
% 1.66/1.88
% 1.66/1.88 This ia a non-Horn set with equality. The strategy will be
% 1.66/1.88 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.66/1.88 deletion, with positive clauses in sos and nonpositive
% 1.66/1.88 clauses in usable.
% 1.66/1.88
% 1.66/1.88 dependent: set(knuth_bendix).
% 1.66/1.88 dependent: set(anl_eq).
% 1.66/1.88 dependent: set(para_from).
% 1.66/1.88 dependent: set(para_into).
% 1.66/1.88 dependent: clear(para_from_right).
% 1.66/1.88 dependent: clear(para_into_right).
% 1.66/1.88 dependent: set(para_from_vars).
% 1.66/1.88 dependent: set(eq_units_both_ways).
% 1.66/1.88 dependent: set(dynamic_demod_all).
% 1.66/1.88 dependent: set(dynamic_demod).
% 1.66/1.88 dependent: set(order_eq).
% 1.66/1.88 dependent: set(back_demod).
% 1.66/1.88 dependent: set(lrpo).
% 1.66/1.88 dependent: set(hyper_res).
% 1.66/1.88 dependent: set(unit_deletion).
% 1.66/1.88 dependent: set(factor).
% 1.66/1.88
% 1.66/1.88 ------------> process usable:
% 1.66/1.88 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.66/1.88 ** KEPT (pick-wt=14): 2 [] A!=unordered_pair(B,C)| -in(D,A)|D=B|D=C.
% 1.66/1.88 ** KEPT (pick-wt=11): 3 [] A!=unordered_pair(B,C)|in(D,A)|D!=B.
% 1.66/1.88 ** KEPT (pick-wt=11): 4 [] A!=unordered_pair(B,C)|in(D,A)|D!=C.
% 1.66/1.88 ** KEPT (pick-wt=17): 5 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=B.
% 1.66/1.88 ** KEPT (pick-wt=17): 6 [] A=unordered_pair(B,C)| -in($f1(B,C,A),A)|$f1(B,C,A)!=C.
% 1.66/1.88 ** KEPT (pick-wt=3): 7 [] $c4!=$c2.
% 1.66/1.88 ** KEPT (pick-wt=3): 8 [] $c4!=$c1.
% 1.66/1.88
% 1.66/1.88 ------------> process sos:
% 1.66/1.88 ** KEPT (pick-wt=3): 11 [] A=A.
% 1.66/1.88 ** KEPT (pick-wt=7): 12 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.66/1.88 ** KEPT (pick-wt=23): 13 [] A=unordered_pair(B,C)|in($f1(B,C,A),A)|$f1(B,C,A)=B|$f1(B,C,A)=C.
% 1.66/1.88 ** KEPT (pick-wt=7): 14 [] unordered_pair($c4,$c3)=unordered_pair($c2,$c1).
% 1.66/1.88 ---> New Demodulator: 15 [new_demod,14] unordered_pair($c4,$c3)=unordered_pair($c2,$c1).
% 1.66/1.88 Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.66/1.88 Following clause subsumed by 12 during input processing: 0 [copy,12,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.66/1.88 >>>> Starting back demodulation with 15.
% 1.66/1.88
% 1.66/1.88 ======= end of input processing =======
% 1.66/1.88
% 1.66/1.88 =========== start of search ===========
% 1.66/1.88
% 1.66/1.88 -------- PROOF --------
% 1.66/1.88
% 1.66/1.88 -----> EMPTY CLAUSE at 0.00 sec ----> 36 [hyper,14,2,18,unit_del,7,8] $F.
% 1.66/1.88
% 1.66/1.88 Length of proof is 1. Level of proof is 1.
% 1.66/1.88
% 1.66/1.88 ---------------- PROOF ----------------
% 1.66/1.88 % SZS status Theorem
% 1.66/1.88 % SZS output start Refutation
% See solution above
% 1.66/1.88 ------------ end of proof -------------
% 1.66/1.88
% 1.66/1.88
% 1.66/1.88 Search stopped by max_proofs option.
% 1.66/1.88
% 1.66/1.88
% 1.66/1.88 Search stopped by max_proofs option.
% 1.66/1.88
% 1.66/1.88 ============ end of search ============
% 1.66/1.88
% 1.66/1.88 -------------- statistics -------------
% 1.66/1.88 clauses given 5
% 1.66/1.88 clauses generated 51
% 1.66/1.88 clauses kept 34
% 1.66/1.88 clauses forward subsumed 30
% 1.66/1.88 clauses back subsumed 0
% 1.66/1.88 Kbytes malloced 976
% 1.66/1.88
% 1.66/1.88 ----------- times (seconds) -----------
% 1.66/1.88 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.88 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.66/1.88 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.66/1.88
% 1.66/1.88 That finishes the proof of the theorem.
% 1.66/1.88
% 1.66/1.88 Process 24775 finished Wed Jul 27 06:56:19 2022
% 1.66/1.88 Otter interrupted
% 1.66/1.88 PROOF FOUND
%------------------------------------------------------------------------------