TSTP Solution File: NUM613+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM613+1 : 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 : n001.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 11:57:16 EDT 2023
% Result : Theorem 12.72s 2.11s
% Output : Proof 12.72s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM613+1 : 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
% 0.12/0.33 % Computer : n001.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Fri Aug 25 14:10:43 EDT 2023
% 0.12/0.33 % CPUTime :
% 12.72/2.11 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 12.72/2.11
% 12.72/2.11 % SZS status Theorem
% 12.72/2.11
% 12.72/2.11 % SZS output start Proof
% 12.72/2.11 Take the following subset of the input axioms:
% 12.72/2.12 fof(mSuccEquSucc, axiom, ![W0, W1]: ((aElementOf0(W0, szNzAzT0) & aElementOf0(W1, szNzAzT0)) => (szszuzczcdt0(W0)=szszuzczcdt0(W1) => W0=W1))).
% 12.72/2.12 fof(m__, conjecture, sbrdtbr0(xP)=xk).
% 12.72/2.12 fof(m__3533, hypothesis, aElementOf0(xk, szNzAzT0) & szszuzczcdt0(xk)=xK).
% 12.72/2.12 fof(m__5255, hypothesis, sbrdtbr0(xQ)=szszuzczcdt0(xk) & (szszuzczcdt0(sbrdtbr0(xP))=sbrdtbr0(xQ) & aElementOf0(sbrdtbr0(xP), szNzAzT0))).
% 12.72/2.12
% 12.72/2.12 Now clausify the problem and encode Horn clauses using encoding 3 of
% 12.72/2.12 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 12.72/2.12 We repeatedly replace C & s=t => u=v by the two clauses:
% 12.72/2.12 fresh(y, y, x1...xn) = u
% 12.72/2.12 C => fresh(s, t, x1...xn) = v
% 12.72/2.12 where fresh is a fresh function symbol and x1..xn are the free
% 12.72/2.12 variables of u and v.
% 12.72/2.12 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 12.72/2.12 input problem has no model of domain size 1).
% 12.72/2.12
% 12.72/2.12 The encoding turns the above axioms into the following unit equations and goals:
% 12.72/2.12
% 12.72/2.12 Axiom 1 (m__3533): szszuzczcdt0(xk) = xK.
% 12.72/2.12 Axiom 2 (m__5255_1): sbrdtbr0(xQ) = szszuzczcdt0(xk).
% 12.72/2.12 Axiom 3 (m__3533_1): aElementOf0(xk, szNzAzT0) = true2.
% 12.72/2.12 Axiom 4 (m__5255): szszuzczcdt0(sbrdtbr0(xP)) = sbrdtbr0(xQ).
% 12.72/2.12 Axiom 5 (m__5255_2): aElementOf0(sbrdtbr0(xP), szNzAzT0) = true2.
% 12.72/2.12 Axiom 6 (mSuccEquSucc): fresh285(X, X, Y, Z) = Z.
% 12.72/2.12 Axiom 7 (mSuccEquSucc): fresh8(X, X, Y, Z) = Y.
% 12.72/2.12 Axiom 8 (mSuccEquSucc): fresh284(X, X, Y, Z) = fresh285(szszuzczcdt0(Y), szszuzczcdt0(Z), Y, Z).
% 12.72/2.12 Axiom 9 (mSuccEquSucc): fresh284(aElementOf0(X, szNzAzT0), true2, Y, X) = fresh8(aElementOf0(Y, szNzAzT0), true2, Y, X).
% 12.72/2.12
% 12.72/2.12 Goal 1 (m__): sbrdtbr0(xP) = xk.
% 12.72/2.12 Proof:
% 12.72/2.12 sbrdtbr0(xP)
% 12.72/2.12 = { by axiom 6 (mSuccEquSucc) R->L }
% 12.72/2.12 fresh285(xK, xK, xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 1 (m__3533) R->L }
% 12.72/2.12 fresh285(xK, szszuzczcdt0(xk), xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 2 (m__5255_1) R->L }
% 12.72/2.12 fresh285(xK, sbrdtbr0(xQ), xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 4 (m__5255) R->L }
% 12.72/2.12 fresh285(xK, szszuzczcdt0(sbrdtbr0(xP)), xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 1 (m__3533) R->L }
% 12.72/2.12 fresh285(szszuzczcdt0(xk), szszuzczcdt0(sbrdtbr0(xP)), xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 8 (mSuccEquSucc) R->L }
% 12.72/2.12 fresh284(true2, true2, xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 5 (m__5255_2) R->L }
% 12.72/2.12 fresh284(aElementOf0(sbrdtbr0(xP), szNzAzT0), true2, xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 9 (mSuccEquSucc) }
% 12.72/2.12 fresh8(aElementOf0(xk, szNzAzT0), true2, xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 3 (m__3533_1) }
% 12.72/2.12 fresh8(true2, true2, xk, sbrdtbr0(xP))
% 12.72/2.12 = { by axiom 7 (mSuccEquSucc) }
% 12.72/2.12 xk
% 12.72/2.12 % SZS output end Proof
% 12.72/2.12
% 12.72/2.12 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------