TSTP Solution File: COL009-1 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : COL009-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n028.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 12:47:59 EDT 2022
% Result : Unsatisfiable 1.65s 1.89s
% Output : Refutation 1.65s
% Verified :
% SZS Type : Refutation
% Derivation depth : 5
% Number of leaves : 3
% Syntax : Number of clauses : 9 ( 9 unt; 0 nHn; 3 RR)
% Number of literals : 9 ( 8 equ; 2 neg)
% Maximal clause size : 1 ( 1 avg)
% Maximal term depth : 7 ( 2 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 4 ( 4 usr; 3 con; 0-2 aty)
% Number of variables : 19 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
A != apply(combinator,A),
file('COL009-1.p',unknown),
[] ).
cnf(2,plain,
apply(combinator,A) != A,
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])]),
[iquote('copy,1,flip.1')] ).
cnf(4,axiom,
apply(apply(apply(b,A),B),C) = apply(A,apply(B,C)),
file('COL009-1.p',unknown),
[] ).
cnf(6,axiom,
apply(apply(l2,A),B) = apply(B,apply(A,A)),
file('COL009-1.p',unknown),
[] ).
cnf(7,plain,
apply(A,apply(B,B)) = apply(apply(l2,B),A),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[6])]),
[iquote('copy,6,flip.1')] ).
cnf(22,plain,
apply(apply(l2,A),apply(apply(b,B),C)) = apply(B,apply(C,apply(A,A))),
inference(para_into,[status(thm),theory(equality)],[4,7]),
[iquote('para_into,4.1.1,7.1.1')] ).
cnf(23,plain,
apply(A,apply(B,apply(C,C))) = apply(apply(l2,C),apply(apply(b,A),B)),
inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[22])]),
[iquote('copy,22,flip.1')] ).
cnf(521,plain,
apply(A,apply(B,apply(C,apply(D,apply(apply(b,C),D))))) = apply(apply(l2,apply(apply(b,C),D)),apply(apply(b,A),B)),
inference(para_into,[status(thm),theory(equality)],[23,4]),
[iquote('para_into,23.1.1.2.2,4.1.1')] ).
cnf(522,plain,
$false,
inference(binary,[status(thm)],[521,2]),
[iquote('binary,521.1,2.1')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : COL009-1 : TPTP v8.1.0. Released v1.0.0.
% 0.06/0.12 % Command : otter-tptp-script %s
% 0.12/0.33 % Computer : n028.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 02:32:48 EDT 2022
% 0.12/0.33 % CPUTime :
% 1.65/1.89 ----- Otter 3.3f, August 2004 -----
% 1.65/1.89 The process was started by sandbox on n028.cluster.edu,
% 1.65/1.89 Wed Jul 27 02:32:48 2022
% 1.65/1.89 The command was "./otter". The process ID is 13234.
% 1.65/1.89
% 1.65/1.89 set(prolog_style_variables).
% 1.65/1.89 set(auto).
% 1.65/1.89 dependent: set(auto1).
% 1.65/1.89 dependent: set(process_input).
% 1.65/1.89 dependent: clear(print_kept).
% 1.65/1.89 dependent: clear(print_new_demod).
% 1.65/1.89 dependent: clear(print_back_demod).
% 1.65/1.89 dependent: clear(print_back_sub).
% 1.65/1.89 dependent: set(control_memory).
% 1.65/1.89 dependent: assign(max_mem, 12000).
% 1.65/1.89 dependent: assign(pick_given_ratio, 4).
% 1.65/1.89 dependent: assign(stats_level, 1).
% 1.65/1.89 dependent: assign(max_seconds, 10800).
% 1.65/1.89 clear(print_given).
% 1.65/1.89
% 1.65/1.89 list(usable).
% 1.65/1.89 0 [] A=A.
% 1.65/1.89 0 [] apply(apply(apply(b,X),Y),Z)=apply(X,apply(Y,Z)).
% 1.65/1.89 0 [] apply(apply(l2,X),Y)=apply(Y,apply(X,X)).
% 1.65/1.89 0 [] Y!=apply(combinator,Y).
% 1.65/1.89 end_of_list.
% 1.65/1.89
% 1.65/1.89 SCAN INPUT: prop=0, horn=1, equality=1, symmetry=0, max_lits=1.
% 1.65/1.89
% 1.65/1.89 All clauses are units, and equality is present; the
% 1.65/1.89 strategy will be Knuth-Bendix with positive clauses in sos.
% 1.65/1.89
% 1.65/1.89 dependent: set(knuth_bendix).
% 1.65/1.89 dependent: set(anl_eq).
% 1.65/1.89 dependent: set(para_from).
% 1.65/1.89 dependent: set(para_into).
% 1.65/1.89 dependent: clear(para_from_right).
% 1.65/1.89 dependent: clear(para_into_right).
% 1.65/1.89 dependent: set(para_from_vars).
% 1.65/1.89 dependent: set(eq_units_both_ways).
% 1.65/1.89 dependent: set(dynamic_demod_all).
% 1.65/1.89 dependent: set(dynamic_demod).
% 1.65/1.89 dependent: set(order_eq).
% 1.65/1.89 dependent: set(back_demod).
% 1.65/1.89 dependent: set(lrpo).
% 1.65/1.89
% 1.65/1.89 ------------> process usable:
% 1.65/1.89 ** KEPT (pick-wt=5): 2 [copy,1,flip.1] apply(combinator,A)!=A.
% 1.65/1.89
% 1.65/1.89 ------------> process sos:
% 1.65/1.89 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.65/1.89 ** KEPT (pick-wt=13): 4 [] apply(apply(apply(b,A),B),C)=apply(A,apply(B,C)).
% 1.65/1.89 ---> New Demodulator: 5 [new_demod,4] apply(apply(apply(b,A),B),C)=apply(A,apply(B,C)).
% 1.65/1.89 ** KEPT (pick-wt=11): 6 [] apply(apply(l2,A),B)=apply(B,apply(A,A)).
% 1.65/1.89 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.65/1.89 >>>> Starting back demodulation with 5.
% 1.65/1.89 ** KEPT (pick-wt=11): 7 [copy,6,flip.1] apply(A,apply(B,B))=apply(apply(l2,B),A).
% 1.65/1.89 Following clause subsumed by 6 during input processing: 0 [copy,7,flip.1] apply(apply(l2,A),B)=apply(B,apply(A,A)).
% 1.65/1.89
% 1.65/1.89 ======= end of input processing =======
% 1.65/1.89
% 1.65/1.89 =========== start of search ===========
% 1.65/1.89
% 1.65/1.89 -------- PROOF --------
% 1.65/1.89
% 1.65/1.89 ----> UNIT CONFLICT at 0.04 sec ----> 522 [binary,521.1,2.1] $F.
% 1.65/1.89
% 1.65/1.89 Length of proof is 5. Level of proof is 4.
% 1.65/1.89
% 1.65/1.89 ---------------- PROOF ----------------
% 1.65/1.89 % SZS status Unsatisfiable
% 1.65/1.89 % SZS output start Refutation
% See solution above
% 1.65/1.89 ------------ end of proof -------------
% 1.65/1.89
% 1.65/1.89
% 1.65/1.89 Search stopped by max_proofs option.
% 1.65/1.89
% 1.65/1.89
% 1.65/1.89 Search stopped by max_proofs option.
% 1.65/1.89
% 1.65/1.89 ============ end of search ============
% 1.65/1.89
% 1.65/1.89 -------------- statistics -------------
% 1.65/1.89 clauses given 27
% 1.65/1.89 clauses generated 569
% 1.65/1.89 clauses kept 466
% 1.65/1.89 clauses forward subsumed 566
% 1.65/1.89 clauses back subsumed 9
% 1.65/1.89 Kbytes malloced 1953
% 1.65/1.89
% 1.65/1.89 ----------- times (seconds) -----------
% 1.65/1.89 user CPU time 0.04 (0 hr, 0 min, 0 sec)
% 1.65/1.89 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.65/1.89 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.65/1.89
% 1.65/1.89 That finishes the proof of the theorem.
% 1.65/1.89
% 1.65/1.89 Process 13234 finished Wed Jul 27 02:32:50 2022
% 1.65/1.89 Otter interrupted
% 1.65/1.89 PROOF FOUND
%------------------------------------------------------------------------------