TSTP Solution File: LCL644+1.001 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : LCL644+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 : n006.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 08:19:41 EDT 2023
% Result : Theorem 0.20s 0.41s
% 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 : LCL644+1.001 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34 % Computer : n006.cluster.edu
% 0.12/0.34 % Model : x86_64 x86_64
% 0.12/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34 % Memory : 8042.1875MB
% 0.12/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34 % CPULimit : 300
% 0.12/0.34 % WCLimit : 300
% 0.12/0.34 % DateTime : Fri Aug 25 02:15:53 EDT 2023
% 0.12/0.34 % CPUTime :
% 0.20/0.41 Command-line arguments: --no-flatten-goal
% 0.20/0.41
% 0.20/0.41 % SZS status Theorem
% 0.20/0.41
% 0.20/0.41 % SZS output start Proof
% 0.20/0.41 Take the following subset of the input axioms:
% 0.20/0.42 fof(main, conjecture, ~?[X]: ~(~(![Y]: (~r1(X, Y) | (~(p2(Y) & ![X2]: (~r1(Y, X2) | p2(X2))) | p1(Y))) | (![Y2]: (~r1(X, Y2) | (p2(Y2) | ~(p1(Y2) & (![X2]: (~r1(Y2, X2) | p1(X2)) & p1(Y2))))) | ~(![Y2]: (~r1(X, Y2) | (![X2]: (~r1(Y2, X2) | (~(p2(X2) & ![Y3]: (~r1(X2, Y3) | p2(Y3))) | p1(X2))) | (p2(Y2) | ~(p1(Y2) & (![X2]: (~r1(Y2, X2) | p1(X2)) & p1(Y2)))))) & (![Y2]: (~r1(X, Y2) | (~(p2(Y2) & ![X2]: (~r1(Y2, X2) | p2(X2))) | (p1(Y2) | ![X2]: (~r1(Y2, X2) | (p2(X2) | ~(p1(X2) & (![Y3]: (~r1(X2, Y3) | p1(Y3)) & p1(X2)))))))) & ![Y2]: (~r1(X, Y2) | (~(p2(Y2) & ![X2]: (~r1(Y2, X2) | p2(X2))) | (p1(Y2) | (p2(Y2) | ~(p1(Y2) & (![X2]: (~r1(Y2, X2) | p1(X2)) & p1(Y2))))))))))) | (![Y2]: (~r1(X, Y2) | (p1(Y2) | ~(![X2]: (~r1(Y2, X2) | p1(X2)) & p1(Y2)))) | ![Y2]: (~r1(X, Y2) | (p1(Y2) | ~(![X2]: (~r1(Y2, X2) | p1(X2)) & p1(Y2))))))).
% 0.20/0.42
% 0.20/0.42 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.42 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.42 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.42 fresh(y, y, x1...xn) = u
% 0.20/0.42 C => fresh(s, t, x1...xn) = v
% 0.20/0.42 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.42 variables of u and v.
% 0.20/0.42 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.42 input problem has no model of domain size 1).
% 0.20/0.42
% 0.20/0.42 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.42
% 0.20/0.42 Axiom 1 (main_2): p1(y2) = true.
% 0.20/0.42 Axiom 2 (main_3): p1(y) = true.
% 0.20/0.42
% 0.20/0.42 Goal 1 (main_23): p1(y) = true.
% 0.20/0.42 Proof:
% 0.20/0.42 p1(y)
% 0.20/0.42 = { by axiom 2 (main_3) }
% 0.20/0.42 true
% 0.20/0.42
% 0.20/0.42 Goal 2 (main_22): p1(y2) = true.
% 0.20/0.42 Proof:
% 0.20/0.42 p1(y2)
% 0.20/0.42 = { by axiom 1 (main_2) }
% 0.20/0.42 true
% 0.20/0.42 % SZS output end Proof
% 0.20/0.42
% 0.20/0.42 RESULT: Theorem (the conjecture is true).
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