TSTP Solution File: LCL186-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : LCL186-1 : 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 : n025.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 08:17:51 EDT 2023
% Result : Unsatisfiable 0.19s 0.52s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : LCL186-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33 % Computer : n025.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Thu Aug 24 23:13:06 EDT 2023
% 0.12/0.33 % CPUTime :
% 0.19/0.52 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.19/0.52
% 0.19/0.52 % SZS status Unsatisfiable
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% 0.19/0.53 % SZS output start Proof
% 0.19/0.53 Take the following subset of the input axioms:
% 0.19/0.53 fof(axiom_1_3, axiom, ![A, B]: axiom(or(not(A), or(B, A)))).
% 0.19/0.53 fof(axiom_1_4, axiom, ![A2, B2]: axiom(or(not(or(A2, B2)), or(B2, A2)))).
% 0.19/0.53 fof(prove_this, negated_conjecture, ~theorem(or(not(not(p)), or(not(p), q)))).
% 0.19/0.53 fof(rule_1, axiom, ![X]: (theorem(X) | ~axiom(X))).
% 0.19/0.53 fof(rule_3, axiom, ![Y, Z, X2]: (theorem(or(not(X2), Z)) | (~axiom(or(not(X2), Y)) | ~theorem(or(not(Y), Z))))).
% 0.19/0.53
% 0.19/0.53 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.53 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.53 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.53 fresh(y, y, x1...xn) = u
% 0.19/0.53 C => fresh(s, t, x1...xn) = v
% 0.19/0.53 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.53 variables of u and v.
% 0.19/0.53 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.53 input problem has no model of domain size 1).
% 0.19/0.53
% 0.19/0.53 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.53
% 0.19/0.53 Axiom 1 (rule_1): fresh4(X, X, Y) = true.
% 0.19/0.53 Axiom 2 (rule_3): fresh(X, X, Y, Z) = true.
% 0.19/0.53 Axiom 3 (rule_1): fresh4(axiom(X), true, X) = theorem(X).
% 0.19/0.53 Axiom 4 (rule_3): fresh2(X, X, Y, Z, W) = theorem(or(not(Y), Z)).
% 0.19/0.53 Axiom 5 (axiom_1_3): axiom(or(not(X), or(Y, X))) = true.
% 0.19/0.53 Axiom 6 (axiom_1_4): axiom(or(not(or(X, Y)), or(Y, X))) = true.
% 0.19/0.53 Axiom 7 (rule_3): fresh2(theorem(or(not(X), Y)), true, Z, Y, X) = fresh(axiom(or(not(Z), X)), true, Z, Y).
% 0.19/0.53
% 0.19/0.53 Goal 1 (prove_this): theorem(or(not(not(p)), or(not(p), q))) = true.
% 0.19/0.53 Proof:
% 0.19/0.53 theorem(or(not(not(p)), or(not(p), q)))
% 0.19/0.53 = { by axiom 4 (rule_3) R->L }
% 0.19/0.53 fresh2(true, true, not(p), or(not(p), q), or(q, not(p)))
% 0.19/0.53 = { by axiom 1 (rule_1) R->L }
% 0.19/0.53 fresh2(fresh4(true, true, or(not(or(q, not(p))), or(not(p), q))), true, not(p), or(not(p), q), or(q, not(p)))
% 0.19/0.53 = { by axiom 6 (axiom_1_4) R->L }
% 0.19/0.53 fresh2(fresh4(axiom(or(not(or(q, not(p))), or(not(p), q))), true, or(not(or(q, not(p))), or(not(p), q))), true, not(p), or(not(p), q), or(q, not(p)))
% 0.19/0.53 = { by axiom 3 (rule_1) }
% 0.19/0.53 fresh2(theorem(or(not(or(q, not(p))), or(not(p), q))), true, not(p), or(not(p), q), or(q, not(p)))
% 0.19/0.53 = { by axiom 7 (rule_3) }
% 0.19/0.53 fresh(axiom(or(not(not(p)), or(q, not(p)))), true, not(p), or(not(p), q))
% 0.19/0.53 = { by axiom 5 (axiom_1_3) }
% 0.19/0.53 fresh(true, true, not(p), or(not(p), q))
% 0.19/0.53 = { by axiom 2 (rule_3) }
% 0.19/0.53 true
% 0.19/0.53 % SZS output end Proof
% 0.19/0.53
% 0.19/0.53 RESULT: Unsatisfiable (the axioms are contradictory).
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