TSTP Solution File: SET917+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SET917+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 13:14:31 EDT 2022
% Result : Theorem 1.91s 2.10s
% Output : Refutation 1.91s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 6
% Syntax : Number of clauses : 10 ( 7 unt; 1 nHn; 7 RR)
% Number of literals : 13 ( 5 equ; 5 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 4 ( 2 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 2 con; 0-2 aty)
% Number of variables : 9 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(2,axiom,
( ~ in(A,B)
| set_intersection2(B,singleton(A)) = singleton(A) ),
file('SET917+1.p',unknown),
[] ).
cnf(5,axiom,
~ disjoint(singleton(dollar_c4),dollar_c3),
file('SET917+1.p',unknown),
[] ).
cnf(6,axiom,
set_intersection2(singleton(dollar_c4),dollar_c3) != singleton(dollar_c4),
file('SET917+1.p',unknown),
[] ).
cnf(8,axiom,
A = A,
file('SET917+1.p',unknown),
[] ).
cnf(9,axiom,
set_intersection2(A,B) = set_intersection2(B,A),
file('SET917+1.p',unknown),
[] ).
cnf(12,axiom,
( in(A,B)
| disjoint(singleton(A),B) ),
file('SET917+1.p',unknown),
[] ).
cnf(14,plain,
( set_intersection2(singleton(A),B) = singleton(A)
| ~ in(A,B) ),
inference(flip,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[9,2])]),
[iquote('para_into,9.1.1,2.2.1,flip.1')] ).
cnf(21,plain,
~ in(dollar_c4,dollar_c3),
inference(unit_del,[status(thm)],[inference(para_from,[status(thm),theory(equality)],[14,6]),8]),
[iquote('para_from,14.1.1,6.1.1,unit_del,8')] ).
cnf(22,plain,
disjoint(singleton(dollar_c4),dollar_c3),
inference(hyper,[status(thm)],[21,12]),
[iquote('hyper,21,12')] ).
cnf(23,plain,
$false,
inference(binary,[status(thm)],[22,5]),
[iquote('binary,22.1,5.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11 % Problem : SET917+1 : TPTP v8.1.0. Released v3.2.0.
% 0.07/0.12 % Command : otter-tptp-script %s
% 0.13/0.33 % Computer : n011.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Wed Jul 27 10:35:40 EDT 2022
% 0.13/0.33 % CPUTime :
% 1.91/2.10 ----- Otter 3.3f, August 2004 -----
% 1.91/2.10 The process was started by sandbox on n011.cluster.edu,
% 1.91/2.10 Wed Jul 27 10:35:40 2022
% 1.91/2.10 The command was "./otter". The process ID is 16495.
% 1.91/2.10
% 1.91/2.10 set(prolog_style_variables).
% 1.91/2.10 set(auto).
% 1.91/2.10 dependent: set(auto1).
% 1.91/2.10 dependent: set(process_input).
% 1.91/2.10 dependent: clear(print_kept).
% 1.91/2.10 dependent: clear(print_new_demod).
% 1.91/2.10 dependent: clear(print_back_demod).
% 1.91/2.10 dependent: clear(print_back_sub).
% 1.91/2.10 dependent: set(control_memory).
% 1.91/2.10 dependent: assign(max_mem, 12000).
% 1.91/2.10 dependent: assign(pick_given_ratio, 4).
% 1.91/2.10 dependent: assign(stats_level, 1).
% 1.91/2.10 dependent: assign(max_seconds, 10800).
% 1.91/2.10 clear(print_given).
% 1.91/2.10
% 1.91/2.10 formula_list(usable).
% 1.91/2.10 all A (A=A).
% 1.91/2.10 all A B (in(A,B)-> -in(B,A)).
% 1.91/2.10 all A B (set_intersection2(A,B)=set_intersection2(B,A)).
% 1.91/2.10 all A B (set_intersection2(A,A)=A).
% 1.91/2.10 all A B (-in(A,B)->disjoint(singleton(A),B)).
% 1.91/2.10 all A B (in(A,B)->set_intersection2(B,singleton(A))=singleton(A)).
% 1.91/2.10 exists A empty(A).
% 1.91/2.10 exists A (-empty(A)).
% 1.91/2.10 all A B (disjoint(A,B)->disjoint(B,A)).
% 1.91/2.10 -(all A B (disjoint(singleton(A),B)|set_intersection2(singleton(A),B)=singleton(A))).
% 1.91/2.10 end_of_list.
% 1.91/2.10
% 1.91/2.10 -------> usable clausifies to:
% 1.91/2.10
% 1.91/2.10 list(usable).
% 1.91/2.10 0 [] A=A.
% 1.91/2.10 0 [] -in(A,B)| -in(B,A).
% 1.91/2.10 0 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.91/2.10 0 [] set_intersection2(A,A)=A.
% 1.91/2.10 0 [] in(A,B)|disjoint(singleton(A),B).
% 1.91/2.10 0 [] -in(A,B)|set_intersection2(B,singleton(A))=singleton(A).
% 1.91/2.10 0 [] empty($c1).
% 1.91/2.10 0 [] -empty($c2).
% 1.91/2.10 0 [] -disjoint(A,B)|disjoint(B,A).
% 1.91/2.10 0 [] -disjoint(singleton($c4),$c3).
% 1.91/2.10 0 [] set_intersection2(singleton($c4),$c3)!=singleton($c4).
% 1.91/2.10 end_of_list.
% 1.91/2.10
% 1.91/2.10 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=2.
% 1.91/2.10
% 1.91/2.10 This ia a non-Horn set with equality. The strategy will be
% 1.91/2.10 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.91/2.10 deletion, with positive clauses in sos and nonpositive
% 1.91/2.10 clauses in usable.
% 1.91/2.10
% 1.91/2.10 dependent: set(knuth_bendix).
% 1.91/2.10 dependent: set(anl_eq).
% 1.91/2.10 dependent: set(para_from).
% 1.91/2.10 dependent: set(para_into).
% 1.91/2.10 dependent: clear(para_from_right).
% 1.91/2.10 dependent: clear(para_into_right).
% 1.91/2.10 dependent: set(para_from_vars).
% 1.91/2.10 dependent: set(eq_units_both_ways).
% 1.91/2.10 dependent: set(dynamic_demod_all).
% 1.91/2.10 dependent: set(dynamic_demod).
% 1.91/2.10 dependent: set(order_eq).
% 1.91/2.10 dependent: set(back_demod).
% 1.91/2.10 dependent: set(lrpo).
% 1.91/2.10 dependent: set(hyper_res).
% 1.91/2.10 dependent: set(unit_deletion).
% 1.91/2.10 dependent: set(factor).
% 1.91/2.10
% 1.91/2.10 ------------> process usable:
% 1.91/2.10 ** KEPT (pick-wt=6): 1 [] -in(A,B)| -in(B,A).
% 1.91/2.10 ** KEPT (pick-wt=10): 2 [] -in(A,B)|set_intersection2(B,singleton(A))=singleton(A).
% 1.91/2.10 ** KEPT (pick-wt=2): 3 [] -empty($c2).
% 1.91/2.10 ** KEPT (pick-wt=6): 4 [] -disjoint(A,B)|disjoint(B,A).
% 1.91/2.10 ** KEPT (pick-wt=4): 5 [] -disjoint(singleton($c4),$c3).
% 1.91/2.10 ** KEPT (pick-wt=7): 6 [] set_intersection2(singleton($c4),$c3)!=singleton($c4).
% 1.91/2.10
% 1.91/2.10 ------------> process sos:
% 1.91/2.10 ** KEPT (pick-wt=3): 8 [] A=A.
% 1.91/2.10 ** KEPT (pick-wt=7): 9 [] set_intersection2(A,B)=set_intersection2(B,A).
% 1.91/2.10 ** KEPT (pick-wt=5): 10 [] set_intersection2(A,A)=A.
% 1.91/2.10 ---> New Demodulator: 11 [new_demod,10] set_intersection2(A,A)=A.
% 1.91/2.10 ** KEPT (pick-wt=7): 12 [] in(A,B)|disjoint(singleton(A),B).
% 1.91/2.10 ** KEPT (pick-wt=2): 13 [] empty($c1).
% 1.91/2.10 Following clause subsumed by 8 during input processing: 0 [copy,8,flip.1] A=A.
% 1.91/2.10 Following clause subsumed by 9 during input processing: 0 [copy,9,flip.1] set_intersection2(A,B)=set_intersection2(B,A).
% 1.91/2.10 >>>> Starting back demodulation with 11.
% 1.91/2.10
% 1.91/2.10 ======= end of input processing =======
% 1.91/2.10
% 1.91/2.10 =========== start of search ===========
% 1.91/2.10
% 1.91/2.10 -------- PROOF --------
% 1.91/2.10
% 1.91/2.10 ----> UNIT CONFLICT at 0.00 sec ----> 23 [binary,22.1,5.1] $F.
% 1.91/2.10
% 1.91/2.10 Length of proof is 3. Level of proof is 3.
% 1.91/2.10
% 1.91/2.10 ---------------- PROOF ----------------
% 1.91/2.10 % SZS status Theorem
% 1.91/2.10 % SZS output start Refutation
% See solution above
% 1.91/2.10 ------------ end of proof -------------
% 1.91/2.10
% 1.91/2.10
% 1.91/2.10 Search stopped by max_proofs option.
% 1.91/2.10
% 1.91/2.10
% 1.91/2.10 Search stopped by max_proofs option.
% 1.91/2.10
% 1.91/2.10 ============ end of search ============
% 1.91/2.10
% 1.91/2.10 -------------- statistics -------------
% 1.91/2.10 clauses given 7
% 1.91/2.10 clauses generated 29
% 1.91/2.10 clauses kept 21
% 1.91/2.10 clauses forward subsumed 21
% 1.91/2.10 clauses back subsumed 0
% 1.91/2.10 Kbytes malloced 976
% 1.91/2.10
% 1.91/2.10 ----------- times (seconds) -----------
% 1.91/2.10 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.91/2.10 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.91/2.10 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.91/2.10
% 1.91/2.10 That finishes the proof of the theorem.
% 1.91/2.10
% 1.91/2.10 Process 16495 finished Wed Jul 27 10:35:42 2022
% 1.91/2.10 Otter interrupted
% 1.91/2.10 PROOF FOUND
%------------------------------------------------------------------------------