TSTP Solution File: SYN386+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : SYN386+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 : 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 : Fri Sep  1 03:34:24 EDT 2023

% Result   : Theorem 0.21s 0.40s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.13  % Problem  : SYN386+1 : TPTP v8.1.2. Released v2.0.0.
% 0.12/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.35  % Computer : n003.cluster.edu
% 0.14/0.35  % Model    : x86_64 x86_64
% 0.14/0.35  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35  % Memory   : 8042.1875MB
% 0.14/0.35  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35  % CPULimit : 300
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Sat Aug 26 17:16:54 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.21/0.40  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.40  
% 0.21/0.40  % SZS status Theorem
% 0.21/0.40  
% 0.21/0.40  % SZS output start Proof
% 0.21/0.40  Take the following subset of the input axioms:
% 0.21/0.41    fof(x2138, conjecture, (![X]: ?[Y]: big_f(X, Y) & (?[X2]: ![E]: ?[N]: ![W]: (big_s(N, W) => big_d(W, X2, E)) & ![E2]: ?[D]: ![B, A2]: (big_d(A2, B, D) => ![Z, Y2]: ((big_f(A2, Y2) & big_f(B, Z)) => big_d(Y2, Z, E2))))) => ?[Y2]: ![E2]: ?[M]: ![W2]: (big_s(M, W2) => ![Z2]: (big_f(W2, Z2) => big_d(Z2, Y2, E2)))).
% 0.21/0.41  
% 0.21/0.41  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.41  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.41  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.41    fresh(y, y, x1...xn) = u
% 0.21/0.41    C => fresh(s, t, x1...xn) = v
% 0.21/0.41  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.41  variables of u and v.
% 0.21/0.41  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.41  input problem has no model of domain size 1).
% 0.21/0.41  
% 0.21/0.41  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.41  
% 0.21/0.41  Axiom 1 (x2138): big_f(X, y(X)) = true2.
% 0.21/0.41  Axiom 2 (x2138_2): big_s(X, w(Y, X)) = true2.
% 0.21/0.41  Axiom 3 (x2138_4): fresh(X, X, Y, Z) = true2.
% 0.21/0.41  Axiom 4 (x2138_3): fresh4(X, X, Y, Z, W) = true2.
% 0.21/0.41  Axiom 5 (x2138_1): big_f(w(X, Y), z(X, Y)) = true2.
% 0.21/0.41  Axiom 6 (x2138_3): fresh2(X, X, Y, Z, W, V) = big_d(W, V, Y).
% 0.21/0.41  Axiom 7 (x2138_4): fresh(big_s(n(X), Y), true2, X, Y) = big_d(Y, x, X).
% 0.21/0.41  Axiom 8 (x2138_3): fresh3(X, X, Y, Z, W, V, U) = fresh4(big_f(Z, V), true2, Y, V, U).
% 0.21/0.41  Axiom 9 (x2138_3): fresh3(big_d(X, Y, d(Z)), true2, Z, X, Y, W, V) = fresh2(big_f(Y, V), true2, Z, X, W, V).
% 0.21/0.41  
% 0.21/0.41  Goal 1 (x2138_5): big_d(z(X, Y), X, e(X)) = true2.
% 0.21/0.41  The goal is true when:
% 0.21/0.41    X = y(x)
% 0.21/0.41    Y = n(d(e(y(x))))
% 0.21/0.41  
% 0.21/0.41  Proof:
% 0.21/0.41    big_d(z(y(x), n(d(e(y(x))))), y(x), e(y(x)))
% 0.21/0.41  = { by axiom 6 (x2138_3) R->L }
% 0.21/0.41    fresh2(true2, true2, e(y(x)), w(y(x), n(d(e(y(x))))), z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 1 (x2138) R->L }
% 0.21/0.41    fresh2(big_f(x, y(x)), true2, e(y(x)), w(y(x), n(d(e(y(x))))), z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 9 (x2138_3) R->L }
% 0.21/0.41    fresh3(big_d(w(y(x), n(d(e(y(x))))), x, d(e(y(x)))), true2, e(y(x)), w(y(x), n(d(e(y(x))))), x, z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 7 (x2138_4) R->L }
% 0.21/0.41    fresh3(fresh(big_s(n(d(e(y(x)))), w(y(x), n(d(e(y(x)))))), true2, d(e(y(x))), w(y(x), n(d(e(y(x)))))), true2, e(y(x)), w(y(x), n(d(e(y(x))))), x, z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 2 (x2138_2) }
% 0.21/0.41    fresh3(fresh(true2, true2, d(e(y(x))), w(y(x), n(d(e(y(x)))))), true2, e(y(x)), w(y(x), n(d(e(y(x))))), x, z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 3 (x2138_4) }
% 0.21/0.41    fresh3(true2, true2, e(y(x)), w(y(x), n(d(e(y(x))))), x, z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 8 (x2138_3) }
% 0.21/0.41    fresh4(big_f(w(y(x), n(d(e(y(x))))), z(y(x), n(d(e(y(x)))))), true2, e(y(x)), z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 5 (x2138_1) }
% 0.21/0.41    fresh4(true2, true2, e(y(x)), z(y(x), n(d(e(y(x))))), y(x))
% 0.21/0.41  = { by axiom 4 (x2138_3) }
% 0.21/0.41    true2
% 0.21/0.41  % SZS output end Proof
% 0.21/0.41  
% 0.21/0.41  RESULT: Theorem (the conjecture is true).
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