%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET883+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n019.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 13:14:28 EDT 2022
% Result : Theorem 1.67s 1.82s
% Output : Refutation 1.67s
% Verified :
% SZS Type : Refutation
% Derivation depth : 2
% Number of leaves : 3
% Syntax : Number of clauses : 5 ( 4 unt; 0 nHn; 5 RR)
% Number of literals : 6 ( 3 equ; 2 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 3 ( 1 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-1 aty)
% Number of variables : 2 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
dollar_c4 != dollar_c3,
file('SET883+1.p',unknown),
[] ).
cnf(3,axiom,
( ~ subset(singleton(A),singleton(B))
| A = B ),
file('SET883+1.p',unknown),
[] ).
cnf(7,axiom,
subset(singleton(dollar_c4),singleton(dollar_c3)),
file('SET883+1.p',unknown),
[] ).
cnf(9,plain,
dollar_c4 = dollar_c3,
inference(hyper,[status(thm)],[7,3]),
[iquote('hyper,7,3')] ).
cnf(11,plain,
$false,
inference(binary,[status(thm)],[9,2]),
[iquote('binary,9.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.03/0.13 % Problem : SET883+1 : TPTP v8.1.0. Released v3.2.0.
% 0.03/0.13 % Command : otter-tptp-script %s
% 0.14/0.34 % Computer : n019.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Wed Jul 27 10:30:37 EDT 2022
% 0.14/0.35 % CPUTime :
% 1.67/1.82 ----- Otter 3.3f, August 2004 -----
% 1.67/1.82 The process was started by sandbox on n019.cluster.edu,
% 1.67/1.82 Wed Jul 27 10:30:37 2022
% 1.67/1.82 The command was "./otter". The process ID is 19683.
% 1.67/1.82
% 1.67/1.82 set(prolog_style_variables).
% 1.67/1.82 set(auto).
% 1.67/1.82 dependent: set(auto1).
% 1.67/1.82 dependent: set(process_input).
% 1.67/1.82 dependent: clear(print_kept).
% 1.67/1.82 dependent: clear(print_new_demod).
% 1.67/1.82 dependent: clear(print_back_demod).
% 1.67/1.82 dependent: clear(print_back_sub).
% 1.67/1.82 dependent: set(control_memory).
% 1.67/1.82 dependent: assign(max_mem, 12000).
% 1.67/1.82 dependent: assign(pick_given_ratio, 4).
% 1.67/1.82 dependent: assign(stats_level, 1).
% 1.67/1.82 dependent: assign(max_seconds, 10800).
% 1.67/1.82 clear(print_given).
% 1.67/1.82
% 1.67/1.82 formula_list(usable).
% 1.67/1.82 all A (A=A).
% 1.67/1.82 all A B subset(A,A).
% 1.67/1.82 exists A empty(A).
% 1.67/1.82 exists A (-empty(A)).
% 1.67/1.82 -(all A B (subset(singleton(A),singleton(B))->A=B)).
% 1.67/1.82 all A B (subset(singleton(A),singleton(B))->A=B).
% 1.67/1.82 end_of_list.
% 1.67/1.82
% 1.67/1.82 -------> usable clausifies to:
% 1.67/1.82
% 1.67/1.82 list(usable).
% 1.67/1.82 0 [] A=A.
% 1.67/1.82 0 [] subset(A,A).
% 1.67/1.82 0 [] empty($c1).
% 1.67/1.82 0 [] -empty($c2).
% 1.67/1.82 0 [] subset(singleton($c4),singleton($c3)).
% 1.67/1.82 0 [] $c4!=$c3.
% 1.67/1.82 0 [] -subset(singleton(A),singleton(B))|A=B.
% 1.67/1.82 end_of_list.
% 1.67/1.82
% 1.67/1.82 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.67/1.82
% 1.67/1.82 This is a Horn set with equality. The strategy will be
% 1.67/1.82 Knuth-Bendix and hyper_res, with positive clauses in
% 1.67/1.82 sos and nonpositive clauses in usable.
% 1.67/1.82
% 1.67/1.82 dependent: set(knuth_bendix).
% 1.67/1.82 dependent: set(anl_eq).
% 1.67/1.82 dependent: set(para_from).
% 1.67/1.82 dependent: set(para_into).
% 1.67/1.82 dependent: clear(para_from_right).
% 1.67/1.82 dependent: clear(para_into_right).
% 1.67/1.82 dependent: set(para_from_vars).
% 1.67/1.82 dependent: set(eq_units_both_ways).
% 1.67/1.82 dependent: set(dynamic_demod_all).
% 1.67/1.82 dependent: set(dynamic_demod).
% 1.67/1.82 dependent: set(order_eq).
% 1.67/1.82 dependent: set(back_demod).
% 1.67/1.82 dependent: set(lrpo).
% 1.67/1.82 dependent: set(hyper_res).
% 1.67/1.82 dependent: clear(order_hyper).
% 1.67/1.82
% 1.67/1.82 ------------> process usable:
% 1.67/1.82 ** KEPT (pick-wt=2): 1 [] -empty($c2).
% 1.67/1.82 ** KEPT (pick-wt=3): 2 [] $c4!=$c3.
% 1.67/1.82 ** KEPT (pick-wt=8): 3 [] -subset(singleton(A),singleton(B))|A=B.
% 1.67/1.82
% 1.67/1.82 ------------> process sos:
% 1.67/1.82 ** KEPT (pick-wt=3): 4 [] A=A.
% 1.67/1.82 ** KEPT (pick-wt=3): 5 [] subset(A,A).
% 1.67/1.82 ** KEPT (pick-wt=2): 6 [] empty($c1).
% 1.67/1.82 ** KEPT (pick-wt=5): 7 [] subset(singleton($c4),singleton($c3)).
% 1.67/1.82 Following clause subsumed by 4 during input processing: 0 [copy,4,flip.1] A=A.
% 1.67/1.82
% 1.67/1.82 ======= end of input processing =======
% 1.67/1.82
% 1.67/1.82 =========== start of search ===========
% 1.67/1.82
% 1.67/1.82 -------- PROOF --------
% 1.67/1.82
% 1.67/1.82 ----> UNIT CONFLICT at 0.00 sec ----> 11 [binary,9.1,2.1] $F.
% 1.67/1.82
% 1.67/1.82 Length of proof is 1. Level of proof is 1.
% 1.67/1.82
% 1.67/1.82 ---------------- PROOF ----------------
% 1.67/1.82 % SZS status Theorem
% 1.67/1.82 % SZS output start Refutation
% See solution above
% 1.67/1.82 ------------ end of proof -------------
% 1.67/1.82
% 1.67/1.82
% 1.67/1.82 Search stopped by max_proofs option.
% 1.67/1.82
% 1.67/1.82
% 1.67/1.82 Search stopped by max_proofs option.
% 1.67/1.82
% 1.67/1.82 ============ end of search ============
% 1.67/1.82
% 1.67/1.82 -------------- statistics -------------
% 1.67/1.82 clauses given 4
% 1.67/1.82 clauses generated 3
% 1.67/1.82 clauses kept 9
% 1.67/1.82 clauses forward subsumed 2
% 1.67/1.82 clauses back subsumed 0
% 1.67/1.82 Kbytes malloced 976
% 1.67/1.82
% 1.67/1.82 ----------- times (seconds) -----------
% 1.67/1.82 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.67/1.82 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.67/1.82 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.67/1.82
% 1.67/1.82 That finishes the proof of the theorem.
% 1.67/1.82
% 1.67/1.82 Process 19683 finished Wed Jul 27 10:30:38 2022
% 1.67/1.82 Otter interrupted
% 1.67/1.82 PROOF FOUND
%------------------------------------------------------------------------------