TSTP Solution File: MSC011+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : MSC011+1 : TPTP v8.1.2. Released v3.2.0.
% 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 09:27:00 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
% 0.00/0.12 % Problem : MSC011+1 : TPTP v8.1.2. Released v3.2.0.
% 0.00/0.13 % 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 : n018.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 : Thu Aug 24 13:54:13 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.13/0.37 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --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(d_cons, axiom, ![A2]: ((drunk(A2) & not_drunk(A2)) => goal)).
% 0.13/0.37 fof(goal_to_be_proved, conjecture, goal).
% 0.13/0.37 fof(neg_phi, axiom, ![A]: (drunk(A) & neg_psi)).
% 0.13/0.37 fof(neg_psi_ax, axiom, neg_psi => ?[A3]: not_drunk(A3)).
% 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 (neg_phi_1): neg_psi = true.
% 0.13/0.37 Axiom 2 (neg_phi): drunk(X) = true.
% 0.13/0.37 Axiom 3 (neg_psi_ax): fresh(X, X) = true.
% 0.13/0.37 Axiom 4 (neg_psi_ax): fresh(neg_psi, true) = not_drunk(a).
% 0.13/0.37 Axiom 5 (d_cons): fresh3(X, X) = true.
% 0.13/0.37 Axiom 6 (d_cons): fresh2(X, X, Y) = goal.
% 0.13/0.37 Axiom 7 (d_cons): fresh2(not_drunk(X), true, X) = fresh3(drunk(X), true).
% 0.13/0.37
% 0.13/0.37 Goal 1 (goal_to_be_proved): goal = true.
% 0.13/0.37 Proof:
% 0.13/0.37 goal
% 0.13/0.37 = { by axiom 6 (d_cons) R->L }
% 0.13/0.37 fresh2(neg_psi, neg_psi, a)
% 0.13/0.37 = { by axiom 1 (neg_phi_1) }
% 0.13/0.37 fresh2(true, neg_psi, a)
% 0.13/0.37 = { by axiom 3 (neg_psi_ax) R->L }
% 0.13/0.37 fresh2(fresh(neg_psi, neg_psi), neg_psi, a)
% 0.13/0.37 = { by axiom 1 (neg_phi_1) }
% 0.13/0.37 fresh2(fresh(neg_psi, true), neg_psi, a)
% 0.13/0.37 = { by axiom 4 (neg_psi_ax) }
% 0.13/0.37 fresh2(not_drunk(a), neg_psi, a)
% 0.13/0.37 = { by axiom 1 (neg_phi_1) }
% 0.13/0.37 fresh2(not_drunk(a), true, a)
% 0.13/0.37 = { by axiom 7 (d_cons) }
% 0.13/0.37 fresh3(drunk(a), true)
% 0.13/0.37 = { by axiom 2 (neg_phi) }
% 0.13/0.37 fresh3(true, true)
% 0.13/0.37 = { by axiom 1 (neg_phi_1) R->L }
% 0.13/0.37 fresh3(neg_psi, true)
% 0.13/0.37 = { by axiom 1 (neg_phi_1) R->L }
% 0.13/0.37 fresh3(neg_psi, neg_psi)
% 0.13/0.37 = { by axiom 5 (d_cons) }
% 0.13/0.37 true
% 0.13/0.37 % SZS output end Proof
% 0.13/0.37
% 0.13/0.37 RESULT: Theorem (the conjecture is true).
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