TSTP Solution File: PUZ011-1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : PUZ011-1 : TPTP v8.1.2. Released v1.0.0.
% 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 : n009.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 13:23:54 EDT 2023

% Result   : Unsatisfiable 0.13s 0.39s
% 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
% 0.06/0.12  % Problem  : PUZ011-1 : TPTP v8.1.2. Released v1.0.0.
% 0.06/0.13  % 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 : n009.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 22:13:05 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.13/0.39  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.13/0.39  
% 0.13/0.39  % SZS status Unsatisfiable
% 0.13/0.39  
% 0.13/0.39  % SZS output start Proof
% 0.13/0.39  Take the following subset of the input axioms:
% 0.13/0.39    fof(india, hypothesis, asian(india)).
% 0.13/0.39    fof(indian, hypothesis, ocean(indian)).
% 0.13/0.39    fof(indian_india, hypothesis, borders(indian, india)).
% 0.13/0.39    fof(indian_somalia, hypothesis, borders(indian, somalia)).
% 0.13/0.39    fof(prove_there_is_a_common_ocean, negated_conjecture, ![Ocean, African, Asian]: (~ocean(Ocean) | (~borders(Ocean, African) | (~african(African) | (~borders(Ocean, Asian) | ~asian(Asian)))))).
% 0.13/0.39    fof(somalia, hypothesis, african(somalia)).
% 0.13/0.39  
% 0.13/0.39  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.13/0.39  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.13/0.39  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.13/0.39    fresh(y, y, x1...xn) = u
% 0.13/0.39    C => fresh(s, t, x1...xn) = v
% 0.13/0.39  where fresh is a fresh function symbol and x1..xn are the free
% 0.13/0.39  variables of u and v.
% 0.13/0.39  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.13/0.39  input problem has no model of domain size 1).
% 0.13/0.39  
% 0.13/0.39  The encoding turns the above axioms into the following unit equations and goals:
% 0.13/0.39  
% 0.13/0.39  Axiom 1 (somalia): african(somalia) = true2.
% 0.13/0.39  Axiom 2 (india): asian(india) = true2.
% 0.13/0.39  Axiom 3 (indian): ocean(indian) = true2.
% 0.13/0.39  Axiom 4 (indian_india): borders(indian, india) = true2.
% 0.13/0.39  Axiom 5 (indian_somalia): borders(indian, somalia) = true2.
% 0.13/0.39  
% 0.13/0.39  Goal 1 (prove_there_is_a_common_ocean): tuple(ocean(X), borders(X, Y), borders(X, Z), african(Y), asian(Z)) = tuple(true2, true2, true2, true2, true2).
% 0.13/0.39  The goal is true when:
% 0.13/0.39    X = indian
% 0.13/0.39    Y = somalia
% 0.13/0.39    Z = india
% 0.13/0.39  
% 0.13/0.39  Proof:
% 0.13/0.39    tuple(ocean(indian), borders(indian, somalia), borders(indian, india), african(somalia), asian(india))
% 0.13/0.39  = { by axiom 4 (indian_india) }
% 0.13/0.39    tuple(ocean(indian), borders(indian, somalia), true2, african(somalia), asian(india))
% 0.13/0.39  = { by axiom 3 (indian) }
% 0.13/0.39    tuple(true2, borders(indian, somalia), true2, african(somalia), asian(india))
% 0.13/0.39  = { by axiom 2 (india) }
% 0.13/0.39    tuple(true2, borders(indian, somalia), true2, african(somalia), true2)
% 0.13/0.39  = { by axiom 1 (somalia) }
% 0.13/0.39    tuple(true2, borders(indian, somalia), true2, true2, true2)
% 0.13/0.39  = { by axiom 5 (indian_somalia) }
% 0.13/0.39    tuple(true2, true2, true2, true2, true2)
% 0.13/0.39  % SZS output end Proof
% 0.13/0.39  
% 0.13/0.39  RESULT: Unsatisfiable (the axioms are contradictory).
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