TSTP Solution File: SYN152-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SYN152-1 : TPTP v8.1.2. Released v1.1.0.
% 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 : n031.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 03:33:22 EDT 2023

% Result   : Unsatisfiable 12.84s 2.12s
% Output   : Proof 12.84s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12  % Problem  : SYN152-1 : TPTP v8.1.2. Released v1.1.0.
% 0.12/0.13  % 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 : n031.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 : Sat Aug 26 21:08:23 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 12.84/2.12  Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 12.84/2.12  
% 12.84/2.12  % SZS status Unsatisfiable
% 12.84/2.12  
% 12.84/2.12  % SZS output start Proof
% 12.84/2.12  Take the following subset of the input axioms:
% 12.84/2.12    fof(axiom_13, axiom, r0(e)).
% 12.84/2.12    fof(axiom_28, axiom, k0(e)).
% 12.84/2.12    fof(prove_this, negated_conjecture, ~n2(e)).
% 12.84/2.12    fof(rule_090, axiom, p1(e, e, e) | (~r0(e) | ~k0(e))).
% 12.84/2.12    fof(rule_137, axiom, ![C, B, A2]: (n2(A2) | ~p1(B, C, A2))).
% 12.84/2.12  
% 12.84/2.12  Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.84/2.12  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.84/2.12  We repeatedly replace C & s=t => u=v by the two clauses:
% 12.84/2.12    fresh(y, y, x1...xn) = u
% 12.84/2.12    C => fresh(s, t, x1...xn) = v
% 12.84/2.12  where fresh is a fresh function symbol and x1..xn are the free
% 12.84/2.12  variables of u and v.
% 12.84/2.12  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.84/2.12  input problem has no model of domain size 1).
% 12.84/2.12  
% 12.84/2.12  The encoding turns the above axioms into the following unit equations and goals:
% 12.84/2.12  
% 12.84/2.12  Axiom 1 (axiom_13): r0(e) = true.
% 12.84/2.12  Axiom 2 (axiom_28): k0(e) = true.
% 12.84/2.12  Axiom 3 (rule_090): fresh320(X, X) = true.
% 12.84/2.12  Axiom 4 (rule_090): fresh321(X, X) = p1(e, e, e).
% 12.84/2.12  Axiom 5 (rule_090): fresh321(k0(e), true) = fresh320(r0(e), true).
% 12.84/2.12  Axiom 6 (rule_137): fresh262(X, X, Y) = true.
% 12.84/2.12  Axiom 7 (rule_137): fresh262(p1(X, Y, Z), true, Z) = n2(Z).
% 12.84/2.12  
% 12.84/2.12  Goal 1 (prove_this): n2(e) = true.
% 12.84/2.12  Proof:
% 12.84/2.12    n2(e)
% 12.84/2.12  = { by axiom 7 (rule_137) R->L }
% 12.84/2.12    fresh262(p1(e, e, e), true, e)
% 12.84/2.12  = { by axiom 4 (rule_090) R->L }
% 12.84/2.12    fresh262(fresh321(true, true), true, e)
% 12.84/2.12  = { by axiom 2 (axiom_28) R->L }
% 12.84/2.12    fresh262(fresh321(k0(e), true), true, e)
% 12.84/2.12  = { by axiom 5 (rule_090) }
% 12.84/2.12    fresh262(fresh320(r0(e), true), true, e)
% 12.84/2.12  = { by axiom 1 (axiom_13) }
% 12.84/2.12    fresh262(fresh320(true, true), true, e)
% 12.84/2.12  = { by axiom 3 (rule_090) }
% 12.84/2.12    fresh262(true, true, e)
% 12.84/2.12  = { by axiom 6 (rule_137) }
% 12.84/2.12    true
% 12.84/2.12  % SZS output end Proof
% 12.84/2.12  
% 12.84/2.12  RESULT: Unsatisfiable (the axioms are contradictory).
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