TSTP Solution File: SWV253-2 by Otter---3.3
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- Process Solution
%------------------------------------------------------------------------------
% File : Otter---3.3
% Problem : SWV253-2 : TPTP v8.1.0. Released v3.2.0.
% Transfm : none
% Format : tptp:raw
% Command : otter-tptp-script %s
% Computer : n020.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 13:20:18 EDT 2022
% Result : Unsatisfiable 1.89s 2.08s
% Output : Refutation 1.89s
% Verified :
% SZS Type : Refutation
% Derivation depth : 3
% Number of leaves : 4
% Syntax : Number of clauses : 8 ( 4 unt; 2 nHn; 6 RR)
% Number of literals : 12 ( 11 equ; 4 neg)
% Maximal clause size : 2 ( 1 avg)
% Maximal term depth : 3 ( 1 avg)
% Number of predicates : 2 ( 0 usr; 1 prp; 0-2 aty)
% Number of functors : 3 ( 3 usr; 2 con; 0-1 aty)
% Number of variables : 2 ( 0 sgn)
% Comments :
%------------------------------------------------------------------------------
cnf(1,axiom,
( v_K != v_K_H
| c_Message_OinvKey(v_K) != c_Message_OinvKey(v_K_H) ),
file('SWV253-2.p',unknown),
[] ).
cnf(2,plain,
( v_K_H != v_K
| c_Message_OinvKey(v_K_H) != c_Message_OinvKey(v_K) ),
inference(flip,[status(thm),theory(equality)],[inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[1])])]),
[iquote('copy,1,flip.1,flip.2')] ).
cnf(3,axiom,
A = A,
file('SWV253-2.p',unknown),
[] ).
cnf(4,axiom,
( c_Message_OinvKey(v_K) = c_Message_OinvKey(v_K_H)
| v_K = v_K_H ),
file('SWV253-2.p',unknown),
[] ).
cnf(5,plain,
( c_Message_OinvKey(v_K_H) = c_Message_OinvKey(v_K)
| v_K_H = v_K ),
inference(flip,[status(thm),theory(equality)],[inference(flip,[status(thm),theory(equality)],[inference(copy,[status(thm)],[4])])]),
[iquote('copy,4,flip.1,flip.2')] ).
cnf(7,axiom,
c_Message_OinvKey(c_Message_OinvKey(A)) = A,
file('SWV253-2.p',unknown),
[] ).
cnf(9,plain,
v_K_H = v_K,
inference(factor_simp,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(para_from,[status(thm),theory(equality)],[5,7]),7])]),
[iquote('para_from,5.1.1,6.1.1.1,demod,7,factor_simp')] ).
cnf(10,plain,
$false,
inference(unit_del,[status(thm)],[inference(demod,[status(thm),theory(equality)],[inference(back_demod,[status(thm)],[2]),9,9]),3,3]),
[iquote('back_demod,2,demod,9,9,unit_del,3,3')] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.12 % Problem : SWV253-2 : TPTP v8.1.0. Released v3.2.0.
% 0.06/0.13 % Command : otter-tptp-script %s
% 0.12/0.34 % Computer : n020.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Wed Jul 27 06:13:38 EDT 2022
% 0.12/0.34 % CPUTime :
% 1.89/2.08 ----- Otter 3.3f, August 2004 -----
% 1.89/2.08 The process was started by sandbox on n020.cluster.edu,
% 1.89/2.08 Wed Jul 27 06:13:38 2022
% 1.89/2.08 The command was "./otter". The process ID is 6043.
% 1.89/2.08
% 1.89/2.08 set(prolog_style_variables).
% 1.89/2.08 set(auto).
% 1.89/2.08 dependent: set(auto1).
% 1.89/2.08 dependent: set(process_input).
% 1.89/2.08 dependent: clear(print_kept).
% 1.89/2.08 dependent: clear(print_new_demod).
% 1.89/2.08 dependent: clear(print_back_demod).
% 1.89/2.08 dependent: clear(print_back_sub).
% 1.89/2.08 dependent: set(control_memory).
% 1.89/2.08 dependent: assign(max_mem, 12000).
% 1.89/2.08 dependent: assign(pick_given_ratio, 4).
% 1.89/2.08 dependent: assign(stats_level, 1).
% 1.89/2.08 dependent: assign(max_seconds, 10800).
% 1.89/2.08 clear(print_given).
% 1.89/2.08
% 1.89/2.08 list(usable).
% 1.89/2.08 0 [] A=A.
% 1.89/2.08 0 [] v_K!=v_K_H|c_Message_OinvKey(v_K)!=c_Message_OinvKey(v_K_H).
% 1.89/2.08 0 [] c_Message_OinvKey(v_K)=c_Message_OinvKey(v_K_H)|v_K=v_K_H.
% 1.89/2.08 0 [] c_Message_OinvKey(c_Message_OinvKey(V_y))=V_y.
% 1.89/2.08 end_of_list.
% 1.89/2.08
% 1.89/2.08 SCAN INPUT: prop=0, horn=0, equality=1, symmetry=0, max_lits=2.
% 1.89/2.08
% 1.89/2.08 This ia a non-Horn set with equality. The strategy will be
% 1.89/2.08 Knuth-Bendix, ordered hyper_res, factoring, and unit
% 1.89/2.08 deletion, with positive clauses in sos and nonpositive
% 1.89/2.08 clauses in usable.
% 1.89/2.08
% 1.89/2.08 dependent: set(knuth_bendix).
% 1.89/2.08 dependent: set(anl_eq).
% 1.89/2.08 dependent: set(para_from).
% 1.89/2.08 dependent: set(para_into).
% 1.89/2.08 dependent: clear(para_from_right).
% 1.89/2.08 dependent: clear(para_into_right).
% 1.89/2.08 dependent: set(para_from_vars).
% 1.89/2.08 dependent: set(eq_units_both_ways).
% 1.89/2.08 dependent: set(dynamic_demod_all).
% 1.89/2.08 dependent: set(dynamic_demod).
% 1.89/2.08 dependent: set(order_eq).
% 1.89/2.08 dependent: set(back_demod).
% 1.89/2.08 dependent: set(lrpo).
% 1.89/2.08 dependent: set(hyper_res).
% 1.89/2.08 dependent: set(unit_deletion).
% 1.89/2.08 dependent: set(factor).
% 1.89/2.08
% 1.89/2.08 ------------> process usable:
% 1.89/2.08 ** KEPT (pick-wt=8): 2 [copy,1,flip.1,flip.2] v_K_H!=v_K|c_Message_OinvKey(v_K_H)!=c_Message_OinvKey(v_K).
% 1.89/2.08
% 1.89/2.08 ------------> process sos:
% 1.89/2.08 ** KEPT (pick-wt=3): 3 [] A=A.
% 1.89/2.08 ** KEPT (pick-wt=8): 5 [copy,4,flip.1,flip.2] c_Message_OinvKey(v_K_H)=c_Message_OinvKey(v_K)|v_K_H=v_K.
% 1.89/2.08 ** KEPT (pick-wt=5): 6 [] c_Message_OinvKey(c_Message_OinvKey(A))=A.
% 1.89/2.08 ---> New Demodulator: 7 [new_demod,6] c_Message_OinvKey(c_Message_OinvKey(A))=A.
% 1.89/2.08 Following clause subsumed by 3 during input processing: 0 [copy,3,flip.1] A=A.
% 1.89/2.08 >>>> Starting back demodulation with 7.
% 1.89/2.08
% 1.89/2.08 ======= end of input processing =======
% 1.89/2.08
% 1.89/2.08 =========== start of search ===========
% 1.89/2.08
% 1.89/2.08 -------- PROOF --------
% 1.89/2.08
% 1.89/2.08 -----> EMPTY CLAUSE at 0.00 sec ----> 10 [back_demod,2,demod,9,9,unit_del,3,3] $F.
% 1.89/2.08
% 1.89/2.08 Length of proof is 3. Level of proof is 2.
% 1.89/2.08
% 1.89/2.08 ---------------- PROOF ----------------
% 1.89/2.08 % SZS status Unsatisfiable
% 1.89/2.08 % SZS output start Refutation
% See solution above
% 1.89/2.08 ------------ end of proof -------------
% 1.89/2.08
% 1.89/2.08
% 1.89/2.08 Search stopped by max_proofs option.
% 1.89/2.08
% 1.89/2.08
% 1.89/2.08 Search stopped by max_proofs option.
% 1.89/2.08
% 1.89/2.08 ============ end of search ============
% 1.89/2.08
% 1.89/2.08 -------------- statistics -------------
% 1.89/2.08 clauses given 3
% 1.89/2.08 clauses generated 16
% 1.89/2.08 clauses kept 5
% 1.89/2.08 clauses forward subsumed 17
% 1.89/2.08 clauses back subsumed 0
% 1.89/2.08 Kbytes malloced 976
% 1.89/2.08
% 1.89/2.08 ----------- times (seconds) -----------
% 1.89/2.08 user CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.89/2.08 system CPU time 0.00 (0 hr, 0 min, 0 sec)
% 1.89/2.08 wall-clock time 2 (0 hr, 0 min, 2 sec)
% 1.89/2.08
% 1.89/2.08 That finishes the proof of the theorem.
% 1.89/2.08
% 1.89/2.08 Process 6043 finished Wed Jul 27 06:13:40 2022
% 1.89/2.08 Otter interrupted
% 1.89/2.08 PROOF FOUND
%------------------------------------------------------------------------------