TSTP Solution File: NUM531+1 by Twee---2.4.2
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% File : Twee---2.4.2
% Problem : NUM531+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 : n025.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 11:56:48 EDT 2023
% Result : Theorem 0.19s 0.38s
% Output : Proof 0.19s
% 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 : NUM531+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
% 0.12/0.34 % Computer : n025.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 15:34:22 EDT 2023
% 0.12/0.34 % CPUTime :
% 0.19/0.38 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.38
% 0.19/0.38 % SZS status Theorem
% 0.19/0.38
% 0.19/0.39 % SZS output start Proof
% 0.19/0.39 Take the following subset of the input axioms:
% 0.19/0.39 fof(mCountNFin, axiom, ![W0]: ((aSet0(W0) & isCountable0(W0)) => ~isFinite0(W0))).
% 0.19/0.39 fof(mDefEmp, definition, ![W0_2]: (W0_2=slcrc0 <=> (aSet0(W0_2) & ~?[W1]: aElementOf0(W1, W0_2)))).
% 0.19/0.39 fof(mEmpFin, axiom, isFinite0(slcrc0)).
% 0.19/0.39 fof(m__, conjecture, ![W0_2]: ((aSet0(W0_2) & isCountable0(W0_2)) => W0_2!=slcrc0)).
% 0.19/0.39
% 0.19/0.39 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.39 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.39 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.39 fresh(y, y, x1...xn) = u
% 0.19/0.39 C => fresh(s, t, x1...xn) = v
% 0.19/0.39 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.39 variables of u and v.
% 0.19/0.39 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.39 input problem has no model of domain size 1).
% 0.19/0.39
% 0.19/0.39 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.39
% 0.19/0.39 Axiom 1 (m__): w0 = slcrc0.
% 0.19/0.39 Axiom 2 (m___1): aSet0(w0) = true2.
% 0.19/0.39 Axiom 3 (mEmpFin): isFinite0(slcrc0) = true2.
% 0.19/0.39 Axiom 4 (m___2): isCountable0(w0) = true2.
% 0.19/0.39
% 0.19/0.39 Goal 1 (mCountNFin): tuple(aSet0(X), isFinite0(X), isCountable0(X)) = tuple(true2, true2, true2).
% 0.19/0.39 The goal is true when:
% 0.19/0.39 X = slcrc0
% 0.19/0.39
% 0.19/0.39 Proof:
% 0.19/0.39 tuple(aSet0(slcrc0), isFinite0(slcrc0), isCountable0(slcrc0))
% 0.19/0.39 = { by axiom 1 (m__) R->L }
% 0.19/0.39 tuple(aSet0(w0), isFinite0(slcrc0), isCountable0(slcrc0))
% 0.19/0.39 = { by axiom 2 (m___1) }
% 0.19/0.39 tuple(true2, isFinite0(slcrc0), isCountable0(slcrc0))
% 0.19/0.39 = { by axiom 3 (mEmpFin) }
% 0.19/0.39 tuple(true2, true2, isCountable0(slcrc0))
% 0.19/0.39 = { by axiom 1 (m__) R->L }
% 0.19/0.39 tuple(true2, true2, isCountable0(w0))
% 0.19/0.39 = { by axiom 4 (m___2) }
% 0.19/0.39 tuple(true2, true2, true2)
% 0.19/0.39 % SZS output end Proof
% 0.19/0.39
% 0.19/0.39 RESULT: Theorem (the conjecture is true).
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