TSTP Solution File: COL060-3 by Otter---3.3
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : COL060-3 : TPTP v8.1.0. Bugfixed v1.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n023.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 12:48:07 EDT 2022
% Result : Unsatisfiable 1.95s 2.11s
% Output : Refutation 1.95s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 4
% Syntax : Number of clauses : 9 ( 9 unt; 0 nHn; 4 RR)
% Number of literals : 9 ( 8 equ; 3 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 9 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 6 ( 6 usr; 5 con; 0-2 aty)
% Number of variables : 11 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
apply(apply(apply(apply(apply(b,apply(apply(b,apply(t,b)),b)),t),x),y),z) != apply(y,apply(x,z)),
file('COL060-3.p',unknown),
[] ).
cnf(2,axiom,
A = A,
file('COL060-3.p',unknown),
[] ).
cnf(4,axiom,
apply(apply(apply(b,A),B),C) = apply(A,apply(B,C)),
file('COL060-3.p',unknown),
[] ).
cnf(5,axiom,
apply(apply(t,A),B) = apply(B,A),
file('COL060-3.p',unknown),
[] ).
cnf(6,plain,
apply(apply(apply(apply(t,b),apply(b,apply(t,x))),y),z) != apply(y,apply(x,z)),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[1]),4,4]),
[iquote('back_demod,1,demod,4,4')] ).
cnf(7,plain,
apply(A,B) = apply(apply(t,B),A),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[5])]),
[iquote('copy,5,flip.1')] ).
cnf(21,plain,
apply(apply(apply(t,A),apply(b,B)),C) = apply(B,apply(A,C)),
inference(para_into,[status(thm),theory(equality)],[4,7]),
[iquote('para_into,3.1.1.1,7.1.1')] ).
cnf(23,plain,
apply(y,apply(x,z)) != apply(y,apply(x,z)),
inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[6]),21,21]),
[iquote('back_demod,6,demod,21,21')] ).
cnf(24,plain,
$false,
inference(binary,[status(thm)],[23,2]),
[iquote('binary,23.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : COL060-3 : TPTP v8.1.0. Bugfixed v1.2.0.
% 0.00/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n023.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Wed Jul 27 02:47:03 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.95/2.11 ----- Otter 3.3f, August 2004 -----
% 1.95/2.11 The process was started by sandbox on n023.cluster.edu,
% 1.95/2.11 Wed Jul 27 02:47:03 2022
% 1.95/2.11 The command was "./otter". The process ID is 2111.
% 1.95/2.11
% 1.95/2.11 set(prolog_style_variables).
% 1.95/2.11 set(auto).
% 1.95/2.11 dependent: set(auto1).
% 1.95/2.11 dependent: set(process_input).
% 1.95/2.11 dependent: clear(print_kept).
% 1.95/2.11 dependent: clear(print_new_demod).
% 1.95/2.11 dependent: clear(print_back_demod).
% 1.95/2.11 dependent: clear(print_back_sub).
% 1.95/2.11 dependent: set(control_memory).
% 1.95/2.11 dependent: assign(max_mem, 12000).
% 1.95/2.11 dependent: assign(pick_given_ratio, 4).
% 1.95/2.11 dependent: assign(stats_level, 1).
% 1.95/2.11 dependent: assign(max_seconds, 10800).
% 1.95/2.11 clear(print_given).
% 1.95/2.11
% 1.95/2.11 list(usable).
% 1.95/2.11 0 [] A=A.
% 1.95/2.11 0 [] apply(apply(apply(b,X),Y),Z)=apply(X,apply(Y,Z)).
% 1.95/2.11 0 [] apply(apply(t,X),Y)=apply(Y,X).
% 1.95/2.11 0 [] apply(apply(apply(apply(apply(b,apply(apply(b,apply(t,b)),b)),t),x),y),z)!=apply(y,apply(x,z)).
% 1.95/2.11 end_of_list.
% 1.95/2.11
% 1.95/2.11 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.95/2.11
% 1.95/2.11 All clauses are units, and equality is present; the
% 1.95/2.11 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.95/2.11
% 1.95/2.11 dependent: set(knuth_bendix).
% 1.95/2.11 dependent: set(anl_eq).
% 1.95/2.11 dependent: set(para_from).
% 1.95/2.11 dependent: set(para_into).
% 1.95/2.11 dependent: clear(para_from_right).
% 1.95/2.11 dependent: clear(para_into_right).
% 1.95/2.11 dependent: set(para_from_vars).
% 1.95/2.11 dependent: set(eq_units_both_ways).
% 1.95/2.11 dependent: set(dynamic_demod_all).
% 1.95/2.11 dependent: set(dynamic_demod).
% 1.95/2.11 dependent: set(order_eq).
% 1.95/2.11 dependent: set(back_demod).
% 1.95/2.11 dependent: set(lrpo).
% 1.95/2.11
% 1.95/2.11 ------------> process usable:
% 1.95/2.11 ** KEPT (pick-wt=23): 1 [] apply(apply(apply(apply(apply(b,apply(apply(b,apply(t,b)),b)),t),x),y),z)!=apply(y,apply(x,z)).
% 1.95/2.11
% 1.95/2.11 ------------> process sos:
% 1.95/2.11 ** KEPT (pick-wt=3): 2 [] A=A.
% 1.95/2.11 ** KEPT (pick-wt=13): 3 [] apply(apply(apply(b,A),B),C)=apply(A,apply(B,C)).
% 1.95/2.11 ---> New Demodulator: 4 [new_demod,3] apply(apply(apply(b,A),B),C)=apply(A,apply(B,C)).
% 1.95/2.11 ** KEPT (pick-wt=9): 5 [] apply(apply(t,A),B)=apply(B,A).
% 1.95/2.11 Following clause subsumed by 2 during input processing: 0 [copy,2,flip.1] A=A.
% 1.95/2.11 >>>> Starting back demodulation with 4.
% 1.95/2.11 >> back demodulating 1 with 4.
% 1.95/2.11 ** KEPT (pick-wt=9): 7 [copy,5,flip.1] apply(A,B)=apply(apply(t,B),A).
% 1.95/2.11 Following clause subsumed by 5 during input processing: 0 [copy,7,flip.1] apply(apply(t,A),B)=apply(B,A).
% 1.95/2.11
% 1.95/2.11 ======= end of input processing =======
% 1.95/2.11
% 1.95/2.11 =========== start of search ===========
% 1.95/2.11
% 1.95/2.11 -------- PROOF --------
% 1.95/2.11
% 1.95/2.11 ----> UNIT CONFLICT at 0.00 sec ----> 24 [binary,23.1,2.1] $F.
% 1.95/2.11
% 1.95/2.11 Length of proof is 4. Level of proof is 3.
% 1.95/2.11
% 1.95/2.11 ---------------- PROOF ----------------
% 1.95/2.11 % SZS status Unsatisfiable
% 1.95/2.11 % SZS output start Refutation
% See solution above
% 1.95/2.11 ------------ end of proof -------------
% 1.95/2.11
% 1.95/2.11
% 1.95/2.11 Search stopped by max_proofs option.
% 1.95/2.11
% 1.95/2.11
% 1.95/2.11 Search stopped by max_proofs option.
% 1.95/2.11
% 1.95/2.11 ============ end of search ============
% 1.95/2.11
% 1.95/2.11 -------------- statistics -------------
% 1.95/2.11 clauses given 6
% 1.95/2.11 clauses generated 26
% 1.95/2.11 clauses kept 17
% 1.95/2.11 clauses forward subsumed 22
% 1.95/2.11 clauses back subsumed 0
% 1.95/2.11 Kbytes malloced 976
% 1.95/2.11
% 1.95/2.11 ----------- times (seconds) -----------
% 1.95/2.11 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.95/2.11 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.95/2.11 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.95/2.11
% 1.95/2.11 That finishes the proof of the theorem.
% 1.95/2.11
% 1.95/2.11 Process 2111 finished Wed Jul 27 02:47:05 2022
% 1.95/2.11 Otter interrupted
% 1.95/2.11 PROOF FOUND
%------------------------------------------------------------------------------