TSTP Solution File: SEU122+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU122+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n006.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 13:14:49 EDT 2022
% Result : Theorem 1.90s 2.09s
% Output : Refutation 1.90s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 6
% Syntax : Number of clauses : 10 ( 7 unt; 1 nHn; 9 RR)
% Number of literals : 14 ( 2 equ; 6 neg)
% Maximal clause size : 3 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 5 ( 3 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 3 con; 0-2 aty)
% Number of variables : 6 ( 1 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(5,axiom,
~ subset(empty_set,dollar_c3),
file('SEU122+1.p',unknown),
[] ).
cnf(7,axiom,
( ~ in(A,B)
| ~ empty(B) ),
file('SEU122+1.p',unknown),
[] ).
cnf(8,axiom,
( ~ empty(A)
| A = B
| ~ empty(B) ),
file('SEU122+1.p',unknown),
[] ).
cnf(12,axiom,
( subset(A,B)
| in(dollar_f1(A,B),A) ),
file('SEU122+1.p',unknown),
[] ).
cnf(13,axiom,
empty(empty_set),
file('SEU122+1.p',unknown),
[] ).
cnf(14,axiom,
empty(dollar_c1),
file('SEU122+1.p',unknown),
[] ).
cnf(17,plain,
empty_set = dollar_c1,
inference(hyper,[status(thm)],[14,8,13]),
[iquote('hyper,14,8,13')] ).
cnf(19,plain,
~ subset(dollar_c1,dollar_c3),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[5]),17]),
[iquote('back_demod,5,demod,17')] ).
cnf(21,plain,
in(dollar_f1(dollar_c1,dollar_c3),dollar_c1),
inference(hyper,[status(thm)],[19,12]),
[iquote('hyper,19,12')] ).
cnf(27,plain,
$false,
inference(hyper,[status(thm)],[21,7,14]),
[iquote('hyper,21,7,14')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.11 % Problem : SEU122+1 : TPTP v8.1.0. Released v3.3.0.
% 0.11/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n006.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 07:55:46 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.90/2.09 ----- Otter 3.3f, August 2004 -----
% 1.90/2.09 The process was started by sandbox on n006.cluster.edu,
% 1.90/2.09 Wed Jul 27 07:55:47 2022
% 1.90/2.09 The command was "./otter". The process ID is 19573.
% 1.90/2.09
% 1.90/2.09 set(prolog_style_variables).
% 1.90/2.09 set(auto).
% 1.90/2.09 dependent: set(auto1).
% 1.90/2.09 dependent: set(process_input).
% 1.90/2.09 dependent: clear(print_kept).
% 1.90/2.09 dependent: clear(print_new_demod).
% 1.90/2.09 dependent: clear(print_back_demod).
% 1.90/2.09 dependent: clear(print_back_sub).
% 1.90/2.09 dependent: set(control_memory).
% 1.90/2.09 dependent: assign(max_mem, 12000).
% 1.90/2.09 dependent: assign(pick_given_ratio, 4).
% 1.90/2.09 dependent: assign(stats_level, 1).
% 1.90/2.09 dependent: assign(max_seconds, 10800).
% 1.90/2.09 clear(print_given).
% 1.90/2.09
% 1.90/2.09 formula_list(usable).
% 1.90/2.09 all A (A=A).
% 1.90/2.09 all A B (in(A,B)-> -in(B,A)).
% 1.90/2.09 all A B (subset(A,B)<-> (all C (in(C,A)->in(C,B)))).
% 1.90/2.09 $T.
% 1.90/2.09 empty(empty_set).
% 1.90/2.09 exists A empty(A).
% 1.90/2.09 exists A (-empty(A)).
% 1.90/2.09 all A B subset(A,A).
% 1.90/2.09 -(all A subset(empty_set,A)).
% 1.90/2.09 all A (empty(A)->A=empty_set).
% 1.90/2.09 all A B (-(in(A,B)&empty(B))).
% 1.90/2.09 all A B (-(empty(A)&A!=B&empty(B))).
% 1.90/2.09 end_of_list.
% 1.90/2.09
% 1.90/2.09 -------> usable clausifies to:
% 1.90/2.09
% 1.90/2.09 list(usable).
% 1.90/2.09 0 [] A=A.
% 1.90/2.09 0 [] -in(A,B)| -in(B,A).
% 1.90/2.09 0 [] -subset(A,B)| -in(C,A)|in(C,B).
% 1.90/2.09 0 [] subset(A,B)|in($f1(A,B),A).
% 1.90/2.09 0 [] subset(A,B)| -in($f1(A,B),B).
% 1.90/2.09 0 [] $T.
% 1.90/2.09 0 [] empty(empty_set).
% 1.90/2.09 0 [] empty($c1).
% 1.90/2.09 0 [] -empty($c2).
% 1.90/2.09 0 [] subset(A,A).
% 1.90/2.09 0 [] -subset(empty_set,$c3).
% 1.90/2.09 0 [] -empty(A)|A=empty_set.
% 1.90/2.09 0 [] -in(A,B)| -empty(B).
% 1.90/2.09 0 [] -empty(A)|A=B| -empty(B).
% 1.90/2.09 end_of_list.
% 1.90/2.09
% 1.90/2.09 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=3.
% 1.90/2.09
% 1.90/2.09 This ia a non-Horn set with equality. The strategy will be
% 1.90/2.09 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.90/2.09 deletion, with positive clauses in sos and nonpositive
% 1.90/2.09 clauses in usable.
% 1.90/2.09
% 1.90/2.09 dependent: set(knuth_bendix).
% 1.90/2.09 dependent: set(anl_eq).
% 1.90/2.09 dependent: set(para_from).
% 1.90/2.09 dependent: set(para_into).
% 1.90/2.09 dependent: clear(para_from_right).
% 1.90/2.09 dependent: clear(para_into_right).
% 1.90/2.09 dependent: set(para_from_vars).
% 1.90/2.09 dependent: set(eq_units_both_ways).
% 1.90/2.09 dependent: set(dynamic_demod_all).
% 1.90/2.09 dependent: set(dynamic_demod).
% 1.90/2.09 dependent: set(order_eq).
% 1.90/2.09 dependent: set(back_demod).
% 1.90/2.09 dependent: set(lrpo).
% 1.90/2.09 dependent: set(hyper_res).
% 1.90/2.09 dependent: set(unit_deletion).
% 1.90/2.09 dependent: set(factor).
% 1.90/2.09
% 1.90/2.09 ------------> process usable:
% 1.90/2.09 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.90/2.09 ** KEPT (pick-wt=9): 2 [] -subset(A,B)| -in(C,A)|in(C,B).
% 1.90/2.09 ** KEPT (pick-wt=8): 3 [] subset(A,B)| -in($f1(A,B),B).
% 1.90/2.09 ** KEPT (pick-wt=2): 4 [] -empty($c2).
% 1.90/2.09 ** KEPT (pick-wt=3): 5 [] -subset(empty_set,$c3).
% 1.90/2.09 ** KEPT (pick-wt=5): 6 [] -empty(A)|A=empty_set.
% 1.90/2.09 ** KEPT (pick-wt=5): 7 [] -in(A,B)| -empty(B).
% 1.90/2.09 ** KEPT (pick-wt=7): 8 [] -empty(A)|A=B| -empty(B).
% 1.90/2.09
% 1.90/2.09 ------------> process sos:
% 1.90/2.09 ** KEPT (pick-wt=3): 11 [] A=A.
% 1.90/2.09 ** KEPT (pick-wt=8): 12 [] subset(A,B)|in($f1(A,B),A).
% 1.90/2.09 ** KEPT (pick-wt=2): 13 [] empty(empty_set).
% 1.90/2.09 ** KEPT (pick-wt=2): 14 [] empty($c1).
% 1.90/2.09 ** KEPT (pick-wt=3): 15 [] subset(A,A).
% 1.90/2.09 Following clause subsumed by 11 during input processing: 0 [copy,11,flip.1] A=A.
% 1.90/2.09 11 back subsumes 10.
% 1.90/2.09
% 1.90/2.09 ======= end of input processing =======
% 1.90/2.09
% 1.90/2.09 =========== start of search ===========
% 1.90/2.09
% 1.90/2.09 -------- PROOF --------
% 1.90/2.09
% 1.90/2.09 -----> EMPTY CLAUSE at 0.00 sec ----> 27 [hyper,21,7,14] $F.
% 1.90/2.09
% 1.90/2.09 Length of proof is 3. Level of proof is 3.
% 1.90/2.09
% 1.90/2.09 ---------------- PROOF ----------------
% 1.90/2.09 % SZS status Theorem
% 1.90/2.09 % SZS output start Refutation
% See solution above
% 1.90/2.09 ------------ end of proof -------------
% 1.90/2.09
% 1.90/2.09
% 1.90/2.09 Search stopped by max_proofs option.
% 1.90/2.09
% 1.90/2.09
% 1.90/2.09 Search stopped by max_proofs option.
% 1.90/2.09
% 1.90/2.09 ============ end of search ============
% 1.90/2.09
% 1.90/2.09 -------------- statistics -------------
% 1.90/2.09 clauses given 10
% 1.90/2.09 clauses generated 33
% 1.90/2.09 clauses kept 25
% 1.90/2.09 clauses forward subsumed 22
% 1.90/2.09 clauses back subsumed 2
% 1.90/2.09 Kbytes malloced 976
% 1.90/2.09
% 1.90/2.09 ----------- times (seconds) -----------
% 1.90/2.09 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.90/2.09 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.90/2.09 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.90/2.09
% 1.90/2.09 That finishes the proof of the theorem.
% 1.90/2.09
% 1.90/2.09 Process 19573 finished Wed Jul 27 07:55:48 2022
% 1.90/2.09 Otter interrupted
% 1.90/2.09 PROOF FOUND
%------------------------------------------------------------------------------