TSTP Solution File: SEU444+1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SEU444+1 : TPTP v8.1.2. Released v3.4.0.
% 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 : 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 : Thu Aug 31 17:52:56 EDT 2023

% Result   : Theorem 0.20s 0.44s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SEU444+1 : TPTP v8.1.2. Released v3.4.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34  % Computer : n020.cluster.edu
% 0.14/0.34  % Model    : x86_64 x86_64
% 0.14/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34  % Memory   : 8042.1875MB
% 0.14/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34  % CPULimit : 300
% 0.14/0.34  % WCLimit  : 300
% 0.14/0.34  % DateTime : Thu Aug 24 00:18:30 EDT 2023
% 0.14/0.34  % CPUTime  : 
% 0.20/0.44  Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.20/0.44  
% 0.20/0.44  % SZS status Theorem
% 0.20/0.44  
% 0.20/0.44  % SZS output start Proof
% 0.20/0.44  Take the following subset of the input axioms:
% 0.20/0.44    fof(t51_relset_2, axiom, ![B, C, A2]: (m2_relset_1(C, B, A2) => (k1_funct_5(C)=k10_relset_1(A2, B, k6_relset_1(B, A2, C), A2) & k2_funct_5(C)=k10_relset_1(B, A2, C, B)))).
% 0.20/0.44    fof(t56_relset_2, conjecture, ![A, B2, C2]: (m2_relset_1(C2, B2, A) => (k1_funct_5(C2)=k10_relset_1(A, B2, k6_relset_1(B2, A, C2), A) & k10_relset_1(A, B2, k6_relset_1(B2, A, C2), k10_relset_1(B2, A, C2, B2))=k10_relset_1(A, B2, k6_relset_1(B2, A, C2), k2_funct_5(C2))))).
% 0.20/0.44  
% 0.20/0.44  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.44  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.44  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.44    fresh(y, y, x1...xn) = u
% 0.20/0.44    C => fresh(s, t, x1...xn) = v
% 0.20/0.44  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.44  variables of u and v.
% 0.20/0.44  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.44  input problem has no model of domain size 1).
% 0.20/0.44  
% 0.20/0.44  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.44  
% 0.20/0.44  Axiom 1 (t56_relset_2): m2_relset_1(c3, b4, a4) = true2.
% 0.20/0.44  Axiom 2 (t51_relset_2): fresh7(X, X, Y, Z, W) = k1_funct_5(W).
% 0.20/0.44  Axiom 3 (t51_relset_2_1): fresh6(X, X, Y, Z, W) = k2_funct_5(W).
% 0.20/0.44  Axiom 4 (t51_relset_2): fresh7(m2_relset_1(X, Y, Z), true2, Z, Y, X) = k10_relset_1(Z, Y, k6_relset_1(Y, Z, X), Z).
% 0.20/0.44  Axiom 5 (t51_relset_2_1): fresh6(m2_relset_1(X, Y, Z), true2, Z, Y, X) = k10_relset_1(Y, Z, X, Y).
% 0.20/0.44  
% 0.20/0.44  Goal 1 (t56_relset_2_1): tuple3(k1_funct_5(c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k10_relset_1(b4, a4, c3, b4))) = tuple3(k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), a4), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k2_funct_5(c3))).
% 0.20/0.44  Proof:
% 0.20/0.44    tuple3(k1_funct_5(c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k10_relset_1(b4, a4, c3, b4)))
% 0.20/0.44  = { by axiom 5 (t51_relset_2_1) R->L }
% 0.20/0.44    tuple3(k1_funct_5(c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), fresh6(m2_relset_1(c3, b4, a4), true2, a4, b4, c3)))
% 0.20/0.44  = { by axiom 1 (t56_relset_2) }
% 0.20/0.44    tuple3(k1_funct_5(c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), fresh6(true2, true2, a4, b4, c3)))
% 0.20/0.44  = { by axiom 3 (t51_relset_2_1) }
% 0.20/0.44    tuple3(k1_funct_5(c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k2_funct_5(c3)))
% 0.20/0.44  = { by axiom 2 (t51_relset_2) R->L }
% 0.20/0.44    tuple3(fresh7(true2, true2, a4, b4, c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k2_funct_5(c3)))
% 0.20/0.44  = { by axiom 1 (t56_relset_2) R->L }
% 0.20/0.44    tuple3(fresh7(m2_relset_1(c3, b4, a4), true2, a4, b4, c3), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k2_funct_5(c3)))
% 0.20/0.44  = { by axiom 4 (t51_relset_2) }
% 0.20/0.44    tuple3(k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), a4), k10_relset_1(a4, b4, k6_relset_1(b4, a4, c3), k2_funct_5(c3)))
% 0.20/0.44  % SZS output end Proof
% 0.20/0.44  
% 0.20/0.44  RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------