TSTP Solution File: SYN379+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN379+1 : TPTP v8.1.2. Released v2.0.0.
% 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 : n007.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 : Fri Sep  1 03:34:22 EDT 2023

% Result   : Theorem 0.13s 0.37s
% Output   : Proof 0.13s
% 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  : SYN379+1 : TPTP v8.1.2. Released v2.0.0.
% 0.07/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34  % Computer : n007.cluster.edu
% 0.13/0.34  % Model    : x86_64 x86_64
% 0.13/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34  % Memory   : 8042.1875MB
% 0.13/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34  % CPULimit : 300
% 0.13/0.34  % WCLimit  : 300
% 0.13/0.34  % DateTime : Sat Aug 26 20:29:57 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.13/0.37  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.13/0.37  
% 0.13/0.37  % SZS status Theorem
% 0.13/0.37  
% 0.13/0.37  % SZS output start Proof
% 0.13/0.37  Take the following subset of the input axioms:
% 0.13/0.37    fof(x2131, conjecture, ![X]: big_p(X) => ?[Y]: (![Z, X2]: big_q(X2, Y, Z) => ~![Z2]: (big_p(Z2) & ~big_q(Y, Y, Z2)))).
% 0.13/0.37  
% 0.13/0.37  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.37  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.37  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.37    fresh(y, y, x1...xn) = u
% 0.13/0.37    C => fresh(s, t, x1...xn) = v
% 0.13/0.37  where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.37  variables of u and v.
% 0.13/0.37  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.37  input problem has no model of domain size 1).
% 0.13/0.37  
% 0.13/0.37  The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.37  
% 0.13/0.37  Axiom 1 (x2131_2): big_q(X, Y, Z) = true2.
% 0.13/0.37  
% 0.13/0.37  Goal 1 (x2131_3): big_q(X, X, Y) = true2.
% 0.13/0.37  The goal is true when:
% 0.13/0.38    X = X
% 0.13/0.38    Y = Y
% 0.13/0.38  
% 0.13/0.38  Proof:
% 0.13/0.38    big_q(X, X, Y)
% 0.13/0.38  = { by axiom 1 (x2131_2) }
% 0.13/0.38    true2
% 0.13/0.38  % SZS output end Proof
% 0.13/0.38  
% 0.13/0.38  RESULT: Theorem (the conjecture is true).
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