TSTP Solution File: HEN002-4 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : HEN002-4 : 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 : n003.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 01:56:51 EDT 2023
% Result : Unsatisfiable 0.20s 0.38s
% 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 : HEN002-4 : TPTP v8.1.2. Released v1.0.0.
% 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 : n003.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 13:31:37 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.20/0.38 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.38
% 0.20/0.38 % SZS status Unsatisfiable
% 0.20/0.38
% 0.20/0.38 % SZS output start Proof
% 0.20/0.38 Take the following subset of the input axioms:
% 0.20/0.38 fof(prove_zero_over_a_is_zero, negated_conjecture, divide(zero, a)!=zero).
% 0.20/0.38 fof(quotient_less_equal1, axiom, ![X, Y]: (~less_equal(X, Y) | divide(X, Y)=zero)).
% 0.20/0.38 fof(zero_is_smallest, axiom, ![X2]: less_equal(zero, X2)).
% 0.20/0.38
% 0.20/0.38 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.38 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.38 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.38 fresh(y, y, x1...xn) = u
% 0.20/0.38 C => fresh(s, t, x1...xn) = v
% 0.20/0.38 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.38 variables of u and v.
% 0.20/0.38 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.38 input problem has no model of domain size 1).
% 0.20/0.38
% 0.20/0.38 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.38
% 0.20/0.38 Axiom 1 (zero_is_smallest): less_equal(zero, X) = true.
% 0.20/0.38 Axiom 2 (quotient_less_equal1): fresh3(X, X, Y, Z) = zero.
% 0.20/0.38 Axiom 3 (quotient_less_equal1): fresh3(less_equal(X, Y), true, X, Y) = divide(X, Y).
% 0.20/0.38
% 0.20/0.38 Goal 1 (prove_zero_over_a_is_zero): divide(zero, a) = zero.
% 0.20/0.38 Proof:
% 0.20/0.38 divide(zero, a)
% 0.20/0.38 = { by axiom 3 (quotient_less_equal1) R->L }
% 0.20/0.38 fresh3(less_equal(zero, a), true, zero, a)
% 0.20/0.38 = { by axiom 1 (zero_is_smallest) }
% 0.20/0.38 fresh3(true, true, zero, a)
% 0.20/0.38 = { by axiom 2 (quotient_less_equal1) }
% 0.20/0.38 zero
% 0.20/0.38 % SZS output end Proof
% 0.20/0.38
% 0.20/0.38 RESULT: Unsatisfiable (the axioms are contradictory).
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