TSTP Solution File: SYN986+1.001 by Twee---2.4.2
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- Process Solution
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% File : Twee---2.4.2
% Problem : SYN986+1.001 : TPTP v8.1.2. Released v4.0.0.
% 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 : n021.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 : Fri Sep 1 03:36:46 EDT 2023
% Result : Theorem 0.20s 0.39s
% 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
% 0.08/0.13 % Problem : SYN986+1.001 : TPTP v8.1.2. Released v4.0.0.
% 0.13/0.14 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.35 % Computer : n021.cluster.edu
% 0.13/0.35 % Model : x86_64 x86_64
% 0.13/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.35 % Memory : 8042.1875MB
% 0.13/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Sat Aug 26 19:37:42 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.39 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.39
% 0.20/0.39 % SZS status Theorem
% 0.20/0.39
% 0.20/0.40 % SZS output start Proof
% 0.20/0.40 Take the following subset of the input axioms:
% 0.20/0.40 fof(ck, conjecture, ?[Z1, Z0]: (r(zero, zero, Z1) & r(zero, Z1, Z0))).
% 0.20/0.40 fof(hyp1, axiom, ![Y]: r(Y, zero, succ(Y))).
% 0.20/0.40 fof(hyp2, axiom, ![X, Z, Y2, Z1_2]: (r(Y2, X, Z) => (r(Z, X, Z1_2) => r(Y2, succ(X), Z1_2)))).
% 0.20/0.40
% 0.20/0.40 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.40 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.40 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.40 fresh(y, y, x1...xn) = u
% 0.20/0.40 C => fresh(s, t, x1...xn) = v
% 0.20/0.40 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.40 variables of u and v.
% 0.20/0.40 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.40 input problem has no model of domain size 1).
% 0.20/0.40
% 0.20/0.40 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.40
% 0.20/0.40 Axiom 1 (hyp1): r(X, zero, succ(X)) = true2.
% 0.20/0.40 Axiom 2 (hyp2): fresh2(X, X, Y, Z, W) = true2.
% 0.20/0.40 Axiom 3 (hyp2): fresh(X, X, Y, Z, W, V) = r(Y, succ(Z), V).
% 0.20/0.40 Axiom 4 (hyp2): fresh(r(X, Y, Z), true2, W, Y, X, Z) = fresh2(r(W, Y, X), true2, W, Y, Z).
% 0.20/0.40
% 0.20/0.40 Goal 1 (ck): tuple(r(zero, X, Y), r(zero, zero, X)) = tuple(true2, true2).
% 0.20/0.40 The goal is true when:
% 0.20/0.40 X = succ(zero)
% 0.20/0.40 Y = succ(succ(zero))
% 0.20/0.40
% 0.20/0.40 Proof:
% 0.20/0.40 tuple(r(zero, succ(zero), succ(succ(zero))), r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 3 (hyp2) R->L }
% 0.20/0.40 tuple(fresh(true2, true2, zero, zero, succ(zero), succ(succ(zero))), r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 1 (hyp1) R->L }
% 0.20/0.40 tuple(fresh(r(succ(zero), zero, succ(succ(zero))), true2, zero, zero, succ(zero), succ(succ(zero))), r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 4 (hyp2) }
% 0.20/0.40 tuple(fresh2(r(zero, zero, succ(zero)), true2, zero, zero, succ(succ(zero))), r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 1 (hyp1) }
% 0.20/0.40 tuple(fresh2(true2, true2, zero, zero, succ(succ(zero))), r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 2 (hyp2) }
% 0.20/0.40 tuple(true2, r(zero, zero, succ(zero)))
% 0.20/0.40 = { by axiom 1 (hyp1) }
% 0.20/0.40 tuple(true2, true2)
% 0.20/0.40 % SZS output end Proof
% 0.20/0.40
% 0.20/0.40 RESULT: Theorem (the conjecture is true).
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