TSTP Solution File: SET891+1 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET891+1 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n011.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 13:14:29 EDT 2022
% Result : Theorem 1.95s 2.14s
% Output : Refutation 1.95s
% Verified :
% SZS Type : Refutation
% Derivation depth : 2
% Number of leaves : 3
% Syntax : Number of clauses : 6 ( 6 unt; 0 nHn; 3 RR)
% Number of literals : 6 ( 5 equ; 2 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 4 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 2 con; 0-2 aty)
% Number of variables : 6 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(4,axiom,
union(unordered_pair(singleton(dollar_c4),singleton(dollar_c3))) != unordered_pair(dollar_c4,dollar_c3),
file('SET891+1.p',unknown),
[] ).
cnf(11,axiom,
union(unordered_pair(A,B)) = set_union2(A,B),
file('SET891+1.p',unknown),
[] ).
cnf(13,axiom,
unordered_pair(A,B) = set_union2(singleton(A),singleton(B)),
file('SET891+1.p',unknown),
[] ).
cnf(14,plain,
set_union2(singleton(dollar_c4),singleton(dollar_c3)) != unordered_pair(dollar_c4,dollar_c3),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[4]),11]),
[iquote('back_demod,4,demod,11')] ).
cnf(15,plain,
set_union2(singleton(A),singleton(B)) = unordered_pair(A,B),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[13])]),
[iquote('copy,13,flip.1')] ).
cnf(16,plain,
$false,
inference(binary,[status(thm)],[15,14]),
[iquote('binary,15.1,14.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SET891+1 : TPTP v8.1.0. Released v3.2.0.
% 0.07/0.12 % Command : otter-tptp-script %s
% 0.12/0.34 % Computer : n011.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Wed Jul 27 10:39:25 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.95/2.14
% 1.95/2.14 -------- PROOF --------
% 1.95/2.14 ----- Otter 3.3f, August 2004 -----
% 1.95/2.14 The process was started by sandbox2 on n011.cluster.edu,
% 1.95/2.14 Wed Jul 27 10:39:26 2022
% 1.95/2.14 The command was "./otter". The process ID is 20991.
% 1.95/2.14
% 1.95/2.14 set(prolog_style_variables).
% 1.95/2.14 set(auto).
% 1.95/2.14 dependent: set(auto1).
% 1.95/2.14 dependent: set(process_input).
% 1.95/2.14 dependent: clear(print_kept).
% 1.95/2.14 dependent: clear(print_new_demod).
% 1.95/2.14 dependent: clear(print_back_demod).
% 1.95/2.14 dependent: clear(print_back_sub).
% 1.95/2.14 dependent: set(control_memory).
% 1.95/2.14 dependent: assign(max_mem, 12000).
% 1.95/2.14 dependent: assign(pick_given_ratio, 4).
% 1.95/2.14 dependent: assign(stats_level, 1).
% 1.95/2.14 dependent: assign(max_seconds, 10800).
% 1.95/2.14 clear(print_given).
% 1.95/2.14
% 1.95/2.14 formula_list(usable).
% 1.95/2.14 all A (A=A).
% 1.95/2.14 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.95/2.14 all A B (set_union2(A,B)=set_union2(B,A)).
% 1.95/2.14 all A B (-empty(A)-> -empty(set_union2(A,B))).
% 1.95/2.14 all A B (-empty(A)-> -empty(set_union2(B,A))).
% 1.95/2.14 all A B (set_union2(A,A)=A).
% 1.95/2.14 all A B (union(unordered_pair(A,B))=set_union2(A,B)).
% 1.95/2.14 exists A empty(A).
% 1.95/2.14 exists A (-empty(A)).
% 1.95/2.14 -(all A B (union(unordered_pair(singleton(A),singleton(B)))=unordered_pair(A,B))).
% 1.95/2.14 all A B (unordered_pair(A,B)=set_union2(singleton(A),singleton(B))).
% 1.95/2.14 end_of_list.
% 1.95/2.14
% 1.95/2.14 -------> usable clausifies to:
% 1.95/2.14
% 1.95/2.14 list(usable).
% 1.95/2.14 0 [] A=A.
% 1.95/2.14 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.95/2.14 0 [] set_union2(A,B)=set_union2(B,A).
% 1.95/2.14 0 [] empty(A)| -empty(set_union2(A,B)).
% 1.95/2.14 0 [] empty(A)| -empty(set_union2(B,A)).
% 1.95/2.14 0 [] set_union2(A,A)=A.
% 1.95/2.14 0 [] union(unordered_pair(A,B))=set_union2(A,B).
% 1.95/2.14 0 [] empty($c1).
% 1.95/2.14 0 [] -empty($c2).
% 1.95/2.14 0 [] union(unordered_pair(singleton($c4),singleton($c3)))!=unordered_pair($c4,$c3).
% 1.95/2.14 0 [] unordered_pair(A,B)=set_union2(singleton(A),singleton(B)).
% 1.95/2.14 end_of_list.
% 1.95/2.14
% 1.95/2.14 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.95/2.14
% 1.95/2.14 This is a Horn set with equality. The strategy will be
% 1.95/2.14 Knuth-Bendix and hyper_res, with positive clauses in
% 1.95/2.14 sos and nonpositive clauses in usable.
% 1.95/2.14
% 1.95/2.14 dependent: set(knuth_bendix).
% 1.95/2.14 dependent: set(anl_eq).
% 1.95/2.14 dependent: set(para_from).
% 1.95/2.14 dependent: set(para_into).
% 1.95/2.14 dependent: clear(para_from_right).
% 1.95/2.14 dependent: clear(para_into_right).
% 1.95/2.14 dependent: set(para_from_vars).
% 1.95/2.14 dependent: set(eq_units_both_ways).
% 1.95/2.14 dependent: set(dynamic_demod_all).
% 1.95/2.14 dependent: set(dynamic_demod).
% 1.95/2.14 dependent: set(order_eq).
% 1.95/2.14 dependent: set(back_demod).
% 1.95/2.14 dependent: set(lrpo).
% 1.95/2.14 dependent: set(hyper_res).
% 1.95/2.14 dependent: clear(order_hyper).
% 1.95/2.14
% 1.95/2.14 ------------> process usable:
% 1.95/2.14 ** KEPT (pick-wt=6): 1 [] empty(A)| -empty(set_union2(A,B)).
% 1.95/2.14 ** KEPT (pick-wt=6): 2 [] empty(A)| -empty(set_union2(B,A)).
% 1.95/2.14 ** KEPT (pick-wt=2): 3 [] -empty($c2).
% 1.95/2.14 ** KEPT (pick-wt=10): 4 [] union(unordered_pair(singleton($c4),singleton($c3)))!=unordered_pair($c4,$c3).
% 1.95/2.14
% 1.95/2.14 ------------> process sos:
% 1.95/2.14 ** KEPT (pick-wt=3): 5 [] A=A.
% 1.95/2.14 ** KEPT (pick-wt=7): 6 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.95/2.14 ** KEPT (pick-wt=7): 7 [] set_union2(A,B)=set_union2(B,A).
% 1.95/2.14 ** KEPT (pick-wt=5): 8 [] set_union2(A,A)=A.
% 1.95/2.14 ---> New Demodulator: 9 [new_demod,8] set_union2(A,A)=A.
% 1.95/2.14 ** KEPT (pick-wt=8): 10 [] union(unordered_pair(A,B))=set_union2(A,B).
% 1.95/2.14 ---> New Demodulator: 11 [new_demod,10] union(unordered_pair(A,B))=set_union2(A,B).
% 1.95/2.14 ** KEPT (pick-wt=2): 12 [] empty($c1).
% 1.95/2.14 ** KEPT (pick-wt=9): 13 [] unordered_pair(A,B)=set_union2(singleton(A),singleton(B)).
% 1.95/2.14 Following clause subsumed by 5 during input processing: 0 [copy,5,flip.1] A=A.
% 1.95/2.14 Following clause subsumed by 6 during input processing: 0 [copy,6,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.95/2.14 Following clause subsumed by 7 during input processing: 0 [copy,7,flip.1] set_union2(A,B)=set_union2(B,A).
% 1.95/2.14 >>>> Starting back demodulation with 9.
% 1.95/2.14 >>>> Starting back demodulation with 11.
% 1.95/2.14 >> back demodulating 4 with 11.
% 1.95/2.14 ** KEPT (pick-wt=9): 15 [copy,13,flip.1] set_union2(singleton(A),singleton(B))=unordered_pair(A,B).
% 1.95/2.14
% 1.95/2.14 ----> UNIT CONFLICT at 0.00 sec ----> 16 [binary,15.1,14.1] $F.
% 1.95/2.14
% 1.95/2.14 Length of proof is 2. Level of proof is 1.
% 1.95/2.14
% 1.95/2.14 ---------------- PROOF ----------------
% 1.95/2.14 % SZS status Theorem
% 1.95/2.14 % SZS output start Refutation
% See solution above
% 1.95/2.14 ------------ end of proof -------------
% 1.95/2.14
% 1.95/2.14
% 1.95/2.14 Search stopped by max_proofs option.
% 1.95/2.14
% 1.95/2.14
% 1.95/2.14 Search stopped by max_proofs option.
% 1.95/2.14
% 1.95/2.14 ============ end of search ============
% 1.95/2.14
% 1.95/2.14 -------------- statistics -------------
% 1.95/2.14 clauses given 0
% 1.95/2.14 clauses generated 0
% 1.95/2.14 clauses kept 13
% 1.95/2.14 clauses forward subsumed 3
% 1.95/2.14 clauses back subsumed 0
% 1.95/2.14 Kbytes malloced 976
% 1.95/2.14
% 1.95/2.14 ----------- times (seconds) -----------
% 1.95/2.14 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.95/2.14 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.95/2.14 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.95/2.14
% 1.95/2.14 That finishes the proof of the theorem.
% 1.95/2.14
% 1.95/2.14 Process 20991 finished Wed Jul 27 10:39:27 2022
% 1.95/2.14 Otter interrupted
% 1.95/2.14 PROOF FOUND
%------------------------------------------------------------------------------