TSTP Solution File: KLE085+1 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : KLE085+1 : TPTP v8.1.0. Released v4.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n027.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 13:00:42 EDT 2022
% Result : Theorem 1.61s 1.84s
% Output : Refutation 1.61s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 5
% Syntax : Number of clauses : 10 ( 10 unt; 0 nHn; 3 RR)
% Number of literals : 10 ( 9 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 : 5 ( 5 usr; 2 con; 0-2 aty)
% Number of variables : 12 ( 1 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(3,axiom,
addition(domain(dollar_c1),one) != one,
file('KLE085+1.p',unknown),
[] ).
cnf(6,axiom,
addition(A,addition(B,C)) = addition(addition(A,B),C),
file('KLE085+1.p',unknown),
[] ).
cnf(7,plain,
addition(addition(A,B),C) = addition(A,addition(B,C)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[6])]),
[iquote('copy,6,flip.1')] ).
cnf(11,axiom,
addition(A,A) = A,
file('KLE085+1.p',unknown),
[] ).
cnf(33,axiom,
addition(antidomain(antidomain(A)),antidomain(A)) = one,
file('KLE085+1.p',unknown),
[] ).
cnf(35,axiom,
domain(A) = antidomain(antidomain(A)),
file('KLE085+1.p',unknown),
[] ).
cnf(44,plain,
addition(antidomain(antidomain(dollar_c1)),one) != one,
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[3]),35]),
[iquote('back_demod,3,demod,35')] ).
cnf(54,plain,
addition(A,addition(A,B)) = addition(A,B),
inference(flip,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[7,11])]),
[iquote('para_into,7.1.1.1,11.1.1,flip.1')] ).
cnf(281,plain,
addition(antidomain(antidomain(A)),one) = one,
inference(demod,[status(thm),theory(equality)],[inference(para_into,[status(thm),theory(equality)],[54,33]),33]),
[iquote('para_into,54.1.1.2,32.1.1,demod,33')] ).
cnf(283,plain,
$false,
inference(binary,[status(thm)],[281,44]),
[iquote('binary,281.1,44.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12 % Problem : KLE085+1 : TPTP v8.1.0. Released v4.0.0.
% 0.04/0.12 % Command : otter-tptp-script %s
% 0.13/0.33 % Computer : n027.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 06:47:21 EDT 2022
% 0.13/0.33 % CPUTime :
% 1.61/1.83 ----- Otter 3.3f, August 2004 -----
% 1.61/1.83 The process was started by sandbox on n027.cluster.edu,
% 1.61/1.83 Wed Jul 27 06:47:21 2022
% 1.61/1.83 The command was "./otter". The process ID is 5777.
% 1.61/1.83
% 1.61/1.83 set(prolog_style_variables).
% 1.61/1.83 set(auto).
% 1.61/1.83 dependent: set(auto1).
% 1.61/1.83 dependent: set(process_input).
% 1.61/1.83 dependent: clear(print_kept).
% 1.61/1.83 dependent: clear(print_new_demod).
% 1.61/1.83 dependent: clear(print_back_demod).
% 1.61/1.83 dependent: clear(print_back_sub).
% 1.61/1.83 dependent: set(control_memory).
% 1.61/1.83 dependent: assign(max_mem, 12000).
% 1.61/1.83 dependent: assign(pick_given_ratio, 4).
% 1.61/1.83 dependent: assign(stats_level, 1).
% 1.61/1.83 dependent: assign(max_seconds, 10800).
% 1.61/1.83 clear(print_given).
% 1.61/1.83
% 1.61/1.83 formula_list(usable).
% 1.61/1.83 all A (A=A).
% 1.61/1.83 all A B (addition(A,B)=addition(B,A)).
% 1.61/1.83 all C B A (addition(A,addition(B,C))=addition(addition(A,B),C)).
% 1.61/1.83 all A (addition(A,zero)=A).
% 1.61/1.83 all A (addition(A,A)=A).
% 1.61/1.83 all A B C (multiplication(A,multiplication(B,C))=multiplication(multiplication(A,B),C)).
% 1.61/1.83 all A (multiplication(A,one)=A).
% 1.61/1.83 all A (multiplication(one,A)=A).
% 1.61/1.83 all A B C (multiplication(A,addition(B,C))=addition(multiplication(A,B),multiplication(A,C))).
% 1.61/1.83 all A B C (multiplication(addition(A,B),C)=addition(multiplication(A,C),multiplication(B,C))).
% 1.61/1.83 all A (multiplication(A,zero)=zero).
% 1.61/1.83 all A (multiplication(zero,A)=zero).
% 1.61/1.83 all A B (le_q(A,B)<->addition(A,B)=B).
% 1.61/1.83 all X0 (multiplication(antidomain(X0),X0)=zero).
% 1.61/1.83 all X0 X1 (addition(antidomain(multiplication(X0,X1)),antidomain(multiplication(X0,antidomain(antidomain(X1)))))=antidomain(multiplication(X0,antidomain(antidomain(X1))))).
% 1.61/1.83 all X0 (addition(antidomain(antidomain(X0)),antidomain(X0))=one).
% 1.61/1.83 all X0 (domain(X0)=antidomain(antidomain(X0))).
% 1.61/1.83 all X0 (multiplication(X0,coantidomain(X0))=zero).
% 1.61/1.83 all X0 X1 (addition(coantidomain(multiplication(X0,X1)),coantidomain(multiplication(coantidomain(coantidomain(X0)),X1)))=coantidomain(multiplication(coantidomain(coantidomain(X0)),X1))).
% 1.61/1.83 all X0 (addition(coantidomain(coantidomain(X0)),coantidomain(X0))=one).
% 1.61/1.83 all X0 (codomain(X0)=coantidomain(coantidomain(X0))).
% 1.61/1.83 -(all X0 (addition(domain(X0),one)=one)).
% 1.61/1.83 end_of_list.
% 1.61/1.83
% 1.61/1.83 -------> usable clausifies to:
% 1.61/1.83
% 1.61/1.83 list(usable).
% 1.61/1.83 0 [] A=A.
% 1.61/1.83 0 [] addition(A,B)=addition(B,A).
% 1.61/1.83 0 [] addition(A,addition(B,C))=addition(addition(A,B),C).
% 1.61/1.83 0 [] addition(A,zero)=A.
% 1.61/1.83 0 [] addition(A,A)=A.
% 1.61/1.83 0 [] multiplication(A,multiplication(B,C))=multiplication(multiplication(A,B),C).
% 1.61/1.83 0 [] multiplication(A,one)=A.
% 1.61/1.83 0 [] multiplication(one,A)=A.
% 1.61/1.83 0 [] multiplication(A,addition(B,C))=addition(multiplication(A,B),multiplication(A,C)).
% 1.61/1.83 0 [] multiplication(addition(A,B),C)=addition(multiplication(A,C),multiplication(B,C)).
% 1.61/1.83 0 [] multiplication(A,zero)=zero.
% 1.61/1.83 0 [] multiplication(zero,A)=zero.
% 1.61/1.83 0 [] -le_q(A,B)|addition(A,B)=B.
% 1.61/1.83 0 [] le_q(A,B)|addition(A,B)!=B.
% 1.61/1.83 0 [] multiplication(antidomain(X0),X0)=zero.
% 1.61/1.83 0 [] addition(antidomain(multiplication(X0,X1)),antidomain(multiplication(X0,antidomain(antidomain(X1)))))=antidomain(multiplication(X0,antidomain(antidomain(X1)))).
% 1.61/1.83 0 [] addition(antidomain(antidomain(X0)),antidomain(X0))=one.
% 1.61/1.83 0 [] domain(X0)=antidomain(antidomain(X0)).
% 1.61/1.83 0 [] multiplication(X0,coantidomain(X0))=zero.
% 1.61/1.83 0 [] addition(coantidomain(multiplication(X0,X1)),coantidomain(multiplication(coantidomain(coantidomain(X0)),X1)))=coantidomain(multiplication(coantidomain(coantidomain(X0)),X1)).
% 1.61/1.83 0 [] addition(coantidomain(coantidomain(X0)),coantidomain(X0))=one.
% 1.61/1.83 0 [] codomain(X0)=coantidomain(coantidomain(X0)).
% 1.61/1.83 0 [] addition(domain($c1),one)!=one.
% 1.61/1.83 end_of_list.
% 1.61/1.83
% 1.61/1.83 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=2.
% 1.61/1.83
% 1.61/1.83 This is a Horn set with equality. The strategy will be
% 1.61/1.83 Knuth-Bendix and hyper_res, with positive clauses in
% 1.61/1.83 sos and nonpositive clauses in usable.
% 1.61/1.83
% 1.61/1.83 dependent: set(knuth_bendix).
% 1.61/1.83 dependent: set(anl_eq).
% 1.61/1.83 dependent: set(para_from).
% 1.61/1.83 dependent: set(para_into).
% 1.61/1.83 dependent: clear(para_from_right).
% 1.61/1.83 dependent: clear(para_into_right).
% 1.61/1.83 dependent: set(para_from_vars).
% 1.61/1.83 dependent: set(eq_units_both_ways).
% 1.61/1.83 dependent: set(dynamic_demod_all).
% 1.61/1.83 dependent: set(dynamic_demod).
% 1.61/1.83 dependent: set(order_eq).
% 1.61/1.83 dependent: set(back_demod).
% 1.61/1.83 dependent: set(lrpo).
% 1.61/1.83 dependent: set(hyper_res).
% 1.61/1.83 dependent: clear(order_hyper).
% 1.61/1.83
% 1.61/1.83 ------------> process usable:
% 1.61/1.83 ** KEPT (pick-wt=8): 1 [] -le_q(A,B)|addition(A,B)=B.
% 1.61/1.83 ** KEPT (pick-wt=8): 2 [] le_q(A,B)|addition(A,B)!=B.
% 1.61/1.83 ** KEPT (pick-wt=6): 3 [] addition(domain($c1),one)!=one.
% 1.61/1.83
% 1.61/1.83 ------------> process sos:
% 1.61/1.83 ** KEPT (pick-wt=3): 4 [] A=A.
% 1.61/1.83 ** KEPT (pick-wt=7): 5 [] addition(A,B)=addition(B,A).
% 1.61/1.83 ** KEPT (pick-wt=11): 7 [copy,6,flip.1] addition(addition(A,B),C)=addition(A,addition(B,C)).
% 1.61/1.83 ---> New Demodulator: 8 [new_demod,7] addition(addition(A,B),C)=addition(A,addition(B,C)).
% 1.61/1.83 ** KEPT (pick-wt=5): 9 [] addition(A,zero)=A.
% 1.61/1.83 ---> New Demodulator: 10 [new_demod,9] addition(A,zero)=A.
% 1.61/1.83 ** KEPT (pick-wt=5): 11 [] addition(A,A)=A.
% 1.61/1.83 ---> New Demodulator: 12 [new_demod,11] addition(A,A)=A.
% 1.61/1.83 ** KEPT (pick-wt=11): 14 [copy,13,flip.1] multiplication(multiplication(A,B),C)=multiplication(A,multiplication(B,C)).
% 1.61/1.83 ---> New Demodulator: 15 [new_demod,14] multiplication(multiplication(A,B),C)=multiplication(A,multiplication(B,C)).
% 1.61/1.83 ** KEPT (pick-wt=5): 16 [] multiplication(A,one)=A.
% 1.61/1.83 ---> New Demodulator: 17 [new_demod,16] multiplication(A,one)=A.
% 1.61/1.83 ** KEPT (pick-wt=5): 18 [] multiplication(one,A)=A.
% 1.61/1.83 ---> New Demodulator: 19 [new_demod,18] multiplication(one,A)=A.
% 1.61/1.83 ** KEPT (pick-wt=13): 20 [] multiplication(A,addition(B,C))=addition(multiplication(A,B),multiplication(A,C)).
% 1.61/1.83 ---> New Demodulator: 21 [new_demod,20] multiplication(A,addition(B,C))=addition(multiplication(A,B),multiplication(A,C)).
% 1.61/1.83 ** KEPT (pick-wt=13): 22 [] multiplication(addition(A,B),C)=addition(multiplication(A,C),multiplication(B,C)).
% 1.61/1.83 ---> New Demodulator: 23 [new_demod,22] multiplication(addition(A,B),C)=addition(multiplication(A,C),multiplication(B,C)).
% 1.61/1.83 ** KEPT (pick-wt=5): 24 [] multiplication(A,zero)=zero.
% 1.61/1.83 ---> New Demodulator: 25 [new_demod,24] multiplication(A,zero)=zero.
% 1.61/1.83 ** KEPT (pick-wt=5): 26 [] multiplication(zero,A)=zero.
% 1.61/1.83 ---> New Demodulator: 27 [new_demod,26] multiplication(zero,A)=zero.
% 1.61/1.83 ** KEPT (pick-wt=6): 28 [] multiplication(antidomain(A),A)=zero.
% 1.61/1.83 ---> New Demodulator: 29 [new_demod,28] multiplication(antidomain(A),A)=zero.
% 1.61/1.83 ** KEPT (pick-wt=18): 30 [] addition(antidomain(multiplication(A,B)),antidomain(multiplication(A,antidomain(antidomain(B)))))=antidomain(multiplication(A,antidomain(antidomain(B)))).
% 1.61/1.83 ---> New Demodulator: 31 [new_demod,30] addition(antidomain(multiplication(A,B)),antidomain(multiplication(A,antidomain(antidomain(B)))))=antidomain(multiplication(A,antidomain(antidomain(B)))).
% 1.61/1.83 ** KEPT (pick-wt=8): 32 [] addition(antidomain(antidomain(A)),antidomain(A))=one.
% 1.61/1.83 ---> New Demodulator: 33 [new_demod,32] addition(antidomain(antidomain(A)),antidomain(A))=one.
% 1.61/1.83 ** KEPT (pick-wt=6): 34 [] domain(A)=antidomain(antidomain(A)).
% 1.61/1.83 ---> New Demodulator: 35 [new_demod,34] domain(A)=antidomain(antidomain(A)).
% 1.61/1.83 ** KEPT (pick-wt=6): 36 [] multiplication(A,coantidomain(A))=zero.
% 1.61/1.83 ---> New Demodulator: 37 [new_demod,36] multiplication(A,coantidomain(A))=zero.
% 1.61/1.83 ** KEPT (pick-wt=18): 38 [] addition(coantidomain(multiplication(A,B)),coantidomain(multiplication(coantidomain(coantidomain(A)),B)))=coantidomain(multiplication(coantidomain(coantidomain(A)),B)).
% 1.61/1.83 ---> New Demodulator: 39 [new_demod,38] addition(coantidomain(multiplication(A,B)),coantidomain(multiplication(coantidomain(coantidomain(A)),B)))=coantidomain(multiplication(coantidomain(coantidomain(A)),B)).
% 1.61/1.83 ** KEPT (pick-wt=8): 40 [] addition(coantidomain(coantidomain(A)),coantidomain(A))=one.
% 1.61/1.83 ---> New Demodulator: 41 [new_demod,40] addition(coantidomain(coantidomain(A)),coantidomain(A))=one.
% 1.61/1.83 ** KEPT (pick-wt=6): 42 [] codomain(A)=coantidomain(coantidomain(A)).
% 1.61/1.83 ---> New Demodulator: 43 [new_demod,42] codomain(A)=coantidomain(coantidomain(A)).
% 1.61/1.83 Following clause subsumed by 4 during input processing: 0 [copy,4,flip.1] A=A.
% 1.61/1.83 Following clause subsumed by 5 during input processing: 0 [copy,5,flip.1] addition(A,B)=addition(B,A).
% 1.61/1.83 >>>> Starting back demodulation with 8.
% 1.61/1.83 >>>> Starting back demodulation with 10.
% 1.61/1.83 >>>> Starting back demodulation with 12.
% 1.61/1.83 >>>> Starting back demodulation with 15.
% 1.61/1.83 >>>> Starting back demodulation with 17.
% 1.61/1.83 >>>> Starting back demodulation with 19.
% 1.61/1.83 >>>> Starting back demodulation with 21.
% 1.61/1.84 >>>> Starting back demodulation with 23.
% 1.61/1.84 >>>> Starting back demodulation with 25.
% 1.61/1.84 >>>> Starting back demodulation with 27.
% 1.61/1.84 >>>> Starting back demodulation with 29.
% 1.61/1.84 >>>> Starting back demodulation with 31.
% 1.61/1.84 >>>> Starting back demodulation with 33.
% 1.61/1.84 >>>> Starting back demodulation with 35.
% 1.61/1.84 >> back demodulating 3 with 35.
% 1.61/1.84 >>>> Starting back demodulation with 37.
% 1.61/1.84 >>>> Starting back demodulation with 39.
% 1.61/1.84 >>>> Starting back demodulation with 41.
% 1.61/1.84 >>>> Starting back demodulation with 43.
% 1.61/1.84
% 1.61/1.84 ======= end of input processing =======
% 1.61/1.84
% 1.61/1.84 =========== start of search ===========
% 1.61/1.84
% 1.61/1.84 -------- PROOF --------
% 1.61/1.84
% 1.61/1.84 ----> UNIT CONFLICT at 0.01 sec ----> 283 [binary,281.1,44.1] $F.
% 1.61/1.84
% 1.61/1.84 Length of proof is 4. Level of proof is 3.
% 1.61/1.84
% 1.61/1.84 ---------------- PROOF ----------------
% 1.61/1.84 % SZS status Theorem
% 1.61/1.84 % SZS output start Refutation
% See solution above
% 1.61/1.84 ------------ end of proof -------------
% 1.61/1.84
% 1.61/1.84
% 1.61/1.84 Search stopped by max_proofs option.
% 1.61/1.84
% 1.61/1.84
% 1.61/1.84 Search stopped by max_proofs option.
% 1.61/1.84
% 1.61/1.84 ============ end of search ============
% 1.61/1.84
% 1.61/1.84 -------------- statistics -------------
% 1.61/1.84 clauses given 46
% 1.61/1.84 clauses generated 586
% 1.61/1.84 clauses kept 204
% 1.61/1.84 clauses forward subsumed 434
% 1.61/1.84 clauses back subsumed 13
% 1.61/1.84 Kbytes malloced 2929
% 1.61/1.84
% 1.61/1.84 ----------- times (seconds) -----------
% 1.61/1.84 user CPU time 0.01 (0 hr, 0 min, 0 sec)
% 1.61/1.84 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.61/1.84 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.61/1.84
% 1.61/1.84 That finishes the proof of the theorem.
% 1.61/1.84
% 1.61/1.84 Process 5777 finished Wed Jul 27 06:47:23 2022
% 1.61/1.84 Otter interrupted
% 1.61/1.84 PROOF FOUND
%------------------------------------------------------------------------------