TSTP Solution File: SWV142+1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWV142+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof

% Computer : n016.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 : Thu Aug 31 23:02:44 EDT 2023

% Result   : Theorem 2.19s 0.68s
% Output   : Proof 2.19s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : SWV142+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35  % Computer : n016.cluster.edu
% 0.13/0.35  % Model    : x86_64 x86_64
% 0.13/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35  % Memory   : 8042.1875MB
% 0.13/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35  % CPULimit : 300
% 0.13/0.35  % WCLimit  : 300
% 0.13/0.35  % DateTime : Tue Aug 29 03:17:30 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 2.19/0.68  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 2.19/0.68  
% 2.19/0.68  % SZS status Theorem
% 2.19/0.68  
% 2.19/0.68  % SZS output start Proof
% 2.19/0.68  Take the following subset of the input axioms:
% 2.19/0.69    fof(gauss_array_0012, conjecture, (leq(tptp_float_0_001, pv1341) & (leq(n1, loopcounter) & ((~leq(tptp_float_0_001, pv1341) => leq(n0, s_best7)) & ((~leq(tptp_float_0_001, pv1341) => leq(n0, s_sworst7)) & ((~leq(tptp_float_0_001, pv1341) => leq(n0, s_worst7)) & ((~leq(tptp_float_0_001, pv1341) => leq(s_best7, n3)) & ((~leq(tptp_float_0_001, pv1341) => leq(s_sworst7, n3)) & ((~leq(tptp_float_0_001, pv1341) => leq(s_worst7, n3)) & ((gt(loopcounter, n0) => leq(n0, s_best7)) & ((gt(loopcounter, n0) => leq(n0, s_sworst7)) & ((gt(loopcounter, n0) => leq(n0, s_worst7)) & ((gt(loopcounter, n0) => leq(s_best7, n3)) & ((gt(loopcounter, n0) => leq(s_sworst7, n3)) & (gt(loopcounter, n0) => leq(s_worst7, n3))))))))))))))) => leq(s_best7, n3)).
% 2.19/0.69    fof(leq_succ_gt, axiom, ![X, Y]: (leq(succ(X), Y) => gt(Y, X))).
% 2.19/0.69    fof(successor_1, axiom, succ(n0)=n1).
% 2.19/0.69  
% 2.19/0.69  Now clausify the problem and encode Horn clauses using encoding 3 of
% 2.19/0.69  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 2.19/0.69  We repeatedly replace C & s=t => u=v by the two clauses:
% 2.19/0.69    fresh(y, y, x1...xn) = u
% 2.19/0.69    C => fresh(s, t, x1...xn) = v
% 2.19/0.69  where fresh is a fresh function symbol and x1..xn are the free
% 2.19/0.69  variables of u and v.
% 2.19/0.69  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 2.19/0.69  input problem has no model of domain size 1).
% 2.19/0.69  
% 2.19/0.69  The encoding turns the above axioms into the following unit equations and goals:
% 2.19/0.69  
% 2.19/0.69  Axiom 1 (successor_1): succ(n0) = n1.
% 2.19/0.69  Axiom 2 (gauss_array_0012_3): leq(n1, loopcounter) = true3.
% 2.19/0.69  Axiom 3 (gauss_array_0012_11): fresh43(X, X) = true3.
% 2.19/0.69  Axiom 4 (gauss_array_0012_11): fresh43(gt(loopcounter, n0), true3) = leq(s_best7, n3).
% 2.19/0.69  Axiom 5 (leq_succ_gt): fresh31(X, X, Y, Z) = true3.
% 2.19/0.69  Axiom 6 (leq_succ_gt): fresh31(leq(succ(X), Y), true3, X, Y) = gt(Y, X).
% 2.19/0.69  
% 2.19/0.69  Goal 1 (gauss_array_0012_14): leq(s_best7, n3) = true3.
% 2.19/0.69  Proof:
% 2.19/0.69    leq(s_best7, n3)
% 2.19/0.69  = { by axiom 4 (gauss_array_0012_11) R->L }
% 2.19/0.69    fresh43(gt(loopcounter, n0), true3)
% 2.19/0.69  = { by axiom 6 (leq_succ_gt) R->L }
% 2.19/0.69    fresh43(fresh31(leq(succ(n0), loopcounter), true3, n0, loopcounter), true3)
% 2.19/0.69  = { by axiom 1 (successor_1) }
% 2.19/0.69    fresh43(fresh31(leq(n1, loopcounter), true3, n0, loopcounter), true3)
% 2.19/0.69  = { by axiom 2 (gauss_array_0012_3) }
% 2.19/0.69    fresh43(fresh31(true3, true3, n0, loopcounter), true3)
% 2.19/0.69  = { by axiom 5 (leq_succ_gt) }
% 2.19/0.69    fresh43(true3, true3)
% 2.19/0.69  = { by axiom 3 (gauss_array_0012_11) }
% 2.19/0.69    true3
% 2.19/0.69  % SZS output end Proof
% 2.19/0.69  
% 2.19/0.69  RESULT: Theorem (the conjecture is true).
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