TSTP Solution File: SEU150+1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SEU150+1 : TPTP v8.1.0. Released v3.3.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n024.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 1.75s 1.93s
% Output : Refutation 1.75s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 4
% Syntax : Number of clauses : 9 ( 7 unt; 0 nHn; 8 RR)
% Number of literals : 11 ( 10 equ; 3 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 2 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 5 ( 5 usr; 3 con; 0-2 aty)
% Number of variables : 7 ( 2 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( singleton(A) != unordered_pair(B,C)
| A = B ),
file('SEU150+1.p',unknown),
[] ).
cnf(2,axiom,
dollar_c2 != dollar_c1,
file('SEU150+1.p',unknown),
[] ).
cnf(4,axiom,
unordered_pair(A,B) = unordered_pair(B,A),
file('SEU150+1.p',unknown),
[] ).
cnf(5,axiom,
singleton(dollar_c3) = unordered_pair(dollar_c2,dollar_c1),
file('SEU150+1.p',unknown),
[] ).
cnf(8,plain,
dollar_c3 = dollar_c2,
inference(hyper,[status(thm)],[5,1]),
[iquote('hyper,5,1')] ).
cnf(10,plain,
singleton(dollar_c2) = unordered_pair(dollar_c2,dollar_c1),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[5]),8]),
[iquote('back_demod,5,demod,8')] ).
cnf(11,plain,
( A = dollar_c2
| unordered_pair(dollar_c2,dollar_c1) != unordered_pair(A,B) ),
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[8,1]),8,10]),
[iquote('para_into,7.1.1,1.2.1,demod,8,10')] ).
cnf(16,plain,
dollar_c2 = dollar_c1,
inference(flip,[status(thm),theory(equality)],[inference(hyper,[status(thm)],[11,4])]),
[iquote('hyper,11,4,flip.1')] ).
cnf(18,plain,
$false,
inference(binary,[status(thm)],[16,2]),
[iquote('binary,16.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SEU150+1 : TPTP v8.1.0. Released v3.3.0.
% 0.07/0.13 % Command : otter-tptp-script %s
% 0.13/0.34 % Computer : n024.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Wed Jul 27 07:45:40 EDT 2022
% 0.13/0.34 % CPUTime :
% 1.75/1.93 ----- Otter 3.3f, August 2004 -----
% 1.75/1.93 The process was started by sandbox on n024.cluster.edu,
% 1.75/1.93 Wed Jul 27 07:45:40 2022
% 1.75/1.93 The command was "./otter". The process ID is 25369.
% 1.75/1.93
% 1.75/1.93 set(prolog_style_variables).
% 1.75/1.93 set(auto).
% 1.75/1.93 dependent: set(auto1).
% 1.75/1.93 dependent: set(process_input).
% 1.75/1.93 dependent: clear(print_kept).
% 1.75/1.93 dependent: clear(print_new_demod).
% 1.75/1.93 dependent: clear(print_back_demod).
% 1.75/1.93 dependent: clear(print_back_sub).
% 1.75/1.93 dependent: set(control_memory).
% 1.75/1.93 dependent: assign(max_mem, 12000).
% 1.75/1.93 dependent: assign(pick_given_ratio, 4).
% 1.75/1.93 dependent: assign(stats_level, 1).
% 1.75/1.93 dependent: assign(max_seconds, 10800).
% 1.75/1.93 clear(print_given).
% 1.75/1.93
% 1.75/1.93 formula_list(usable).
% 1.75/1.93 all A (A=A).
% 1.75/1.93 all A B (unordered_pair(A,B)=unordered_pair(B,A)).
% 1.75/1.93 $T.
% 1.75/1.93 $T.
% 1.75/1.93 all A B C (singleton(A)=unordered_pair(B,C)->A=B).
% 1.75/1.93 -(all A B C (singleton(A)=unordered_pair(B,C)->B=C)).
% 1.75/1.93 end_of_list.
% 1.75/1.93
% 1.75/1.93 -------> usable clausifies to:
% 1.75/1.93
% 1.75/1.93 list(usable).
% 1.75/1.93 0 [] A=A.
% 1.75/1.93 0 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.93 0 [] $T.
% 1.75/1.93 0 [] $T.
% 1.75/1.93 0 [] singleton(A)!=unordered_pair(B,C)|A=B.
% 1.75/1.93 0 [] singleton($c3)=unordered_pair($c2,$c1).
% 1.75/1.93 0 [] $c2!=$c1.
% 1.75/1.93 end_of_list.
% 1.75/1.93
% 1.75/1.93 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.75/1.93
% 1.75/1.93 This is a Horn set with equality. The strategy will be
% 1.75/1.93 Knuth-Bendix and hyper_res, with positive clauses in
% 1.75/1.93 sos and nonpositive clauses in usable.
% 1.75/1.93
% 1.75/1.93 dependent: set(knuth_bendix).
% 1.75/1.93 dependent: set(anl_eq).
% 1.75/1.93 dependent: set(para_from).
% 1.75/1.93 dependent: set(para_into).
% 1.75/1.93 dependent: clear(para_from_right).
% 1.75/1.93 dependent: clear(para_into_right).
% 1.75/1.93 dependent: set(para_from_vars).
% 1.75/1.93 dependent: set(eq_units_both_ways).
% 1.75/1.93 dependent: set(dynamic_demod_all).
% 1.75/1.93 dependent: set(dynamic_demod).
% 1.75/1.93 dependent: set(order_eq).
% 1.75/1.93 dependent: set(back_demod).
% 1.75/1.93 dependent: set(lrpo).
% 1.75/1.93 dependent: set(hyper_res).
% 1.75/1.93 dependent: clear(order_hyper).
% 1.75/1.93
% 1.75/1.93 ------------> process usable:
% 1.75/1.93 ** KEPT (pick-wt=9): 1 [] singleton(A)!=unordered_pair(B,C)|A=B.
% 1.75/1.93 ** KEPT (pick-wt=3): 2 [] $c2!=$c1.
% 1.75/1.93
% 1.75/1.93 ------------> process sos:
% 1.75/1.93 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.75/1.93 ** KEPT (pick-wt=7): 4 [] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.93 ** KEPT (pick-wt=6): 5 [] singleton($c3)=unordered_pair($c2,$c1).
% 1.75/1.93 ---> New Demodulator: 6 [new_demod,5] singleton($c3)=unordered_pair($c2,$c1).
% 1.75/1.93 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.75/1.93 Following clause subsumed by 4 during input processing: 0 [copy,4,flip.1] unordered_pair(A,B)=unordered_pair(B,A).
% 1.75/1.93 >>>> Starting back demodulation with 6.
% 1.75/1.93
% 1.75/1.93 ======= end of input processing =======
% 1.75/1.93
% 1.75/1.93 =========== start of search ===========
% 1.75/1.93
% 1.75/1.93 -------- PROOF --------
% 1.75/1.93
% 1.75/1.93 ----> UNIT CONFLICT at 0.00 sec ----> 18 [binary,16.1,2.1] $F.
% 1.75/1.93
% 1.75/1.93 Length of proof is 4. Level of proof is 4.
% 1.75/1.93
% 1.75/1.93 ---------------- PROOF ----------------
% 1.75/1.93 % SZS status Theorem
% 1.75/1.93 % SZS output start Refutation
% See solution above
% 1.75/1.93 ------------ end of proof -------------
% 1.75/1.93
% 1.75/1.93
% 1.75/1.93 Search stopped by max_proofs option.
% 1.75/1.93
% 1.75/1.93
% 1.75/1.93 Search stopped by max_proofs option.
% 1.75/1.93
% 1.75/1.93 ============ end of search ============
% 1.75/1.93
% 1.75/1.93 -------------- statistics -------------
% 1.75/1.93 clauses given 6
% 1.75/1.93 clauses generated 14
% 1.75/1.93 clauses kept 13
% 1.75/1.93 clauses forward subsumed 9
% 1.75/1.93 clauses back subsumed 0
% 1.75/1.93 Kbytes malloced 976
% 1.75/1.93
% 1.75/1.93 ----------- times (seconds) -----------
% 1.75/1.93 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.93 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.75/1.93 wall-clock time 1 (0 hr, 0 min, 1 sec)
% 1.75/1.93
% 1.75/1.93 That finishes the proof of the theorem.
% 1.75/1.93
% 1.75/1.93 Process 25369 finished Wed Jul 27 07:45:41 2022
% 1.75/1.93 Otter interrupted
% 1.75/1.93 PROOF FOUND
%------------------------------------------------------------------------------