TSTP Solution File: PLA020-1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : PLA020-1 : 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 : n008.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 13:03:16 EDT 2023
% Result : Unsatisfiable 0.20s 0.43s
% Output : Proof 0.20s
% 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 : PLA020-1 : TPTP v8.1.2. Released v1.1.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.34 % Computer : n008.cluster.edu
% 0.15/0.34 % Model : x86_64 x86_64
% 0.15/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.34 % Memory : 8042.1875MB
% 0.15/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.34 % CPULimit : 300
% 0.15/0.34 % WCLimit : 300
% 0.15/0.34 % DateTime : Sun Aug 27 06:09:47 EDT 2023
% 0.15/0.34 % CPUTime :
% 0.20/0.43 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
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% 0.20/0.43 % SZS status Unsatisfiable
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% 0.20/0.43 % SZS output start Proof
% 0.20/0.43 Take the following subset of the input axioms:
% 0.20/0.43 fof(initial_state3, axiom, holds(on(c, d), s0)).
% 0.20/0.43 fof(initial_state7, axiom, holds(clear(c), s0)).
% 0.20/0.43 fof(initial_state8, axiom, holds(empty, s0)).
% 0.20/0.43 fof(pickup_2, axiom, ![X, Y, State]: (holds(clear(Y), do(pickup(X), State)) | (~holds(on(X, Y), State) | (~holds(clear(X), State) | ~holds(empty, State))))).
% 0.20/0.43 fof(prove_D, negated_conjecture, ![State2]: ~holds(clear(d), State2)).
% 0.20/0.43
% 0.20/0.43 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.43 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.43 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.43 fresh(y, y, x1...xn) = u
% 0.20/0.43 C => fresh(s, t, x1...xn) = v
% 0.20/0.43 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.43 variables of u and v.
% 0.20/0.43 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.43 input problem has no model of domain size 1).
% 0.20/0.43
% 0.20/0.43 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.43
% 0.20/0.43 Axiom 1 (initial_state8): holds(empty, s0) = true2.
% 0.20/0.43 Axiom 2 (initial_state7): holds(clear(c), s0) = true2.
% 0.20/0.43 Axiom 3 (initial_state3): holds(on(c, d), s0) = true2.
% 0.20/0.43 Axiom 4 (pickup_2): fresh20(X, X, Y, Z, W) = true2.
% 0.20/0.43 Axiom 5 (pickup_2): fresh15(X, X, Y, Z, W) = holds(clear(Y), do(pickup(Z), W)).
% 0.20/0.43 Axiom 6 (pickup_2): fresh19(X, X, Y, Z, W) = fresh20(holds(empty, W), true2, Y, Z, W).
% 0.20/0.44 Axiom 7 (pickup_2): fresh19(holds(on(X, Y), Z), true2, Y, X, Z) = fresh15(holds(clear(X), Z), true2, Y, X, Z).
% 0.20/0.44
% 0.20/0.44 Goal 1 (prove_D): holds(clear(d), X) = true2.
% 0.20/0.44 The goal is true when:
% 0.20/0.44 X = do(pickup(c), s0)
% 0.20/0.44
% 0.20/0.44 Proof:
% 0.20/0.44 holds(clear(d), do(pickup(c), s0))
% 0.20/0.44 = { by axiom 5 (pickup_2) R->L }
% 0.20/0.44 fresh15(true2, true2, d, c, s0)
% 0.20/0.44 = { by axiom 2 (initial_state7) R->L }
% 0.20/0.44 fresh15(holds(clear(c), s0), true2, d, c, s0)
% 0.20/0.44 = { by axiom 7 (pickup_2) R->L }
% 0.20/0.44 fresh19(holds(on(c, d), s0), true2, d, c, s0)
% 0.20/0.44 = { by axiom 3 (initial_state3) }
% 0.20/0.44 fresh19(true2, true2, d, c, s0)
% 0.20/0.44 = { by axiom 6 (pickup_2) }
% 0.20/0.44 fresh20(holds(empty, s0), true2, d, c, s0)
% 0.20/0.44 = { by axiom 1 (initial_state8) }
% 0.20/0.44 fresh20(true2, true2, d, c, s0)
% 0.20/0.44 = { by axiom 4 (pickup_2) }
% 0.20/0.44 true2
% 0.20/0.44 % SZS output end Proof
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% 0.20/0.44 RESULT: Unsatisfiable (the axioms are contradictory).
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