TSTP Solution File: SWC291-1 by Twee---2.4.2

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWC291-1 : TPTP v8.1.2. Released v2.4.0.
% 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 : n018.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 20:54:49 EDT 2023

% Result   : Unsatisfiable 2.90s 0.80s
% Output   : Proof 2.90s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : SWC291-1 : TPTP v8.1.2. Released v2.4.0.
% 0.00/0.14  % 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 : n018.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 : Mon Aug 28 14:57:32 EDT 2023
% 0.13/0.35  % CPUTime  : 
% 2.90/0.80  Command-line arguments: --ground-connectedness --complete-subsets
% 2.90/0.80  
% 2.90/0.80  % SZS status Unsatisfiable
% 2.90/0.80  
% 2.90/0.80  % SZS output start Proof
% 2.90/0.80  Take the following subset of the input axioms:
% 2.90/0.80    fof(co1_10, negated_conjecture, ~strictorderedP(sk1)).
% 2.90/0.80    fof(co1_6, negated_conjecture, sk1=sk3).
% 2.90/0.80    fof(co1_8, negated_conjecture, strictorderedP(sk3)).
% 2.90/0.80  
% 2.90/0.80  Now clausify the problem and encode Horn clauses using encoding 3 of
% 2.90/0.80  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 2.90/0.80  We repeatedly replace C & s=t => u=v by the two clauses:
% 2.90/0.80    fresh(y, y, x1...xn) = u
% 2.90/0.80    C => fresh(s, t, x1...xn) = v
% 2.90/0.80  where fresh is a fresh function symbol and x1..xn are the free
% 2.90/0.80  variables of u and v.
% 2.90/0.80  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 2.90/0.80  input problem has no model of domain size 1).
% 2.90/0.80  
% 2.90/0.80  The encoding turns the above axioms into the following unit equations and goals:
% 2.90/0.80  
% 2.90/0.80  Axiom 1 (co1_6): sk1 = sk3.
% 2.90/0.80  Axiom 2 (co1_8): strictorderedP(sk3) = true2.
% 2.90/0.80  
% 2.90/0.80  Goal 1 (co1_10): strictorderedP(sk1) = true2.
% 2.90/0.80  Proof:
% 2.90/0.80    strictorderedP(sk1)
% 2.90/0.80  = { by axiom 1 (co1_6) }
% 2.90/0.80    strictorderedP(sk3)
% 2.90/0.80  = { by axiom 2 (co1_8) }
% 2.90/0.80    true2
% 2.90/0.80  % SZS output end Proof
% 2.90/0.80  
% 2.90/0.80  RESULT: Unsatisfiable (the axioms are contradictory).
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