TSTP Solution File: SYN065+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : SYN065+1 : TPTP v8.1.2. Released v2.0.0.
% 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 : n016.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 03:32:59 EDT 2023
% Result : Theorem 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 : SYN065+1 : TPTP v8.1.2. Released v2.0.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n016.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Sat Aug 26 22:24:42 EDT 2023
% 0.14/0.34 % CPUTime :
% 0.20/0.38 Command-line arguments: --no-flatten-goal
% 0.20/0.38
% 0.20/0.38 % SZS status Theorem
% 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(pel36, conjecture, ![X]: ?[Y]: big_h(X, Y)).
% 0.20/0.38 fof(pel36_1, axiom, ![X2]: ?[Y2]: big_f(X2, Y2)).
% 0.20/0.38 fof(pel36_2, axiom, ![X2]: ?[Y2]: big_g(X2, Y2)).
% 0.20/0.39 fof(pel36_3, axiom, ![X2, Y2]: ((big_f(X2, Y2) | big_g(X2, Y2)) => ![Z]: ((big_f(Y2, Z) | big_g(Y2, Z)) => big_h(X2, Z)))).
% 0.20/0.39
% 0.20/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.39 fresh(y, y, x1...xn) = u
% 0.20/0.39 C => fresh(s, t, x1...xn) = v
% 0.20/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.39 variables of u and v.
% 0.20/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.39 input problem has no model of domain size 1).
% 0.20/0.39
% 0.20/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.39
% 0.20/0.39 Axiom 1 (pel36_1): big_f(X, y2(X)) = true2.
% 0.20/0.39 Axiom 2 (pel36_2): big_g(X, y(X)) = true2.
% 0.20/0.39 Axiom 3 (pel36_3_2): fresh3(X, X, Y, Z) = true2.
% 0.20/0.39 Axiom 4 (pel36_3_2): fresh4(X, X, Y, Z, W) = big_h(Y, W).
% 0.20/0.39 Axiom 5 (pel36_3_2): fresh4(big_g(X, Y), true2, X, Y, Z) = fresh3(big_f(Y, Z), true2, X, Z).
% 0.20/0.39
% 0.20/0.39 Goal 1 (pel36): big_h(x, X) = true2.
% 0.20/0.39 The goal is true when:
% 0.20/0.39 X = y2(y(x))
% 0.20/0.39
% 0.20/0.39 Proof:
% 0.20/0.39 big_h(x, y2(y(x)))
% 0.20/0.39 = { by axiom 4 (pel36_3_2) R->L }
% 0.20/0.39 fresh4(true2, true2, x, y(x), y2(y(x)))
% 0.20/0.39 = { by axiom 2 (pel36_2) R->L }
% 0.20/0.39 fresh4(big_g(x, y(x)), true2, x, y(x), y2(y(x)))
% 0.20/0.39 = { by axiom 5 (pel36_3_2) }
% 0.20/0.39 fresh3(big_f(y(x), y2(y(x))), true2, x, y2(y(x)))
% 0.20/0.39 = { by axiom 1 (pel36_1) }
% 0.20/0.39 fresh3(true2, true2, x, y2(y(x)))
% 0.20/0.39 = { by axiom 3 (pel36_3_2) }
% 0.20/0.39 true2
% 0.20/0.39 % SZS output end Proof
% 0.20/0.39
% 0.20/0.39 RESULT: Theorem (the conjecture is true).
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