TSTP Solution File: NUM606+3 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM606+3 : 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 : n002.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 11:57:14 EDT 2023
% Result : Theorem 9.15s 1.57s
% Output : Proof 9.15s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12 % Problem : NUM606+3 : TPTP v8.1.2. Released v4.0.0.
% 0.12/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n002.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Fri Aug 25 17:08:32 EDT 2023
% 0.13/0.34 % CPUTime :
% 9.15/1.57 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 9.15/1.57
% 9.15/1.57 % SZS status Theorem
% 9.15/1.57
% 9.15/1.58 % SZS output start Proof
% 9.15/1.58 Take the following subset of the input axioms:
% 9.15/1.58 fof(m__, conjecture, ![W0]: (aElementOf0(W0, xQ) => aElementOf0(W0, szNzAzT0)) | aSubsetOf0(xQ, szNzAzT0)).
% 9.15/1.58 fof(m__3435, hypothesis, aSet0(xS) & (![W0_2]: (aElementOf0(W0_2, xS) => aElementOf0(W0_2, szNzAzT0)) & (aSubsetOf0(xS, szNzAzT0) & isCountable0(xS)))).
% 9.15/1.58 fof(m__4998, hypothesis, ![W0_2]: (aElementOf0(W0_2, xO) => aElementOf0(W0_2, xS)) & aSubsetOf0(xO, xS)).
% 9.15/1.58 fof(m__5078, hypothesis, aSet0(xQ) & (![W0_2]: (aElementOf0(W0_2, xQ) => aElementOf0(W0_2, xO)) & (aSubsetOf0(xQ, xO) & (sbrdtbr0(xQ)=xK & aElementOf0(xQ, slbdtsldtrb0(xO, xK)))))).
% 9.15/1.58
% 9.15/1.58 Now clausify the problem and encode Horn clauses using encoding 3 of
% 9.15/1.58 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 9.15/1.58 We repeatedly replace C & s=t => u=v by the two clauses:
% 9.15/1.58 fresh(y, y, x1...xn) = u
% 9.15/1.58 C => fresh(s, t, x1...xn) = v
% 9.15/1.58 where fresh is a fresh function symbol and x1..xn are the free
% 9.15/1.58 variables of u and v.
% 9.15/1.58 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 9.15/1.58 input problem has no model of domain size 1).
% 9.15/1.58
% 9.15/1.58 The encoding turns the above axioms into the following unit equations and goals:
% 9.15/1.58
% 9.15/1.58 Axiom 1 (m__): aElementOf0(w0, xQ) = true2.
% 9.15/1.58 Axiom 2 (m__3435_3): fresh155(X, X, Y) = true2.
% 9.15/1.58 Axiom 3 (m__4998_1): fresh18(X, X, Y) = true2.
% 9.15/1.58 Axiom 4 (m__5078_4): fresh17(X, X, Y) = true2.
% 9.15/1.58 Axiom 5 (m__3435_3): fresh155(aElementOf0(X, xS), true2, X) = aElementOf0(X, szNzAzT0).
% 9.15/1.58 Axiom 6 (m__4998_1): fresh18(aElementOf0(X, xO), true2, X) = aElementOf0(X, xS).
% 9.15/1.58 Axiom 7 (m__5078_4): fresh17(aElementOf0(X, xQ), true2, X) = aElementOf0(X, xO).
% 9.15/1.58
% 9.15/1.58 Goal 1 (m___1): aElementOf0(w0, szNzAzT0) = true2.
% 9.15/1.58 Proof:
% 9.15/1.58 aElementOf0(w0, szNzAzT0)
% 9.15/1.58 = { by axiom 5 (m__3435_3) R->L }
% 9.15/1.58 fresh155(aElementOf0(w0, xS), true2, w0)
% 9.15/1.58 = { by axiom 6 (m__4998_1) R->L }
% 9.15/1.58 fresh155(fresh18(aElementOf0(w0, xO), true2, w0), true2, w0)
% 9.15/1.58 = { by axiom 7 (m__5078_4) R->L }
% 9.15/1.58 fresh155(fresh18(fresh17(aElementOf0(w0, xQ), true2, w0), true2, w0), true2, w0)
% 9.15/1.58 = { by axiom 1 (m__) }
% 9.15/1.58 fresh155(fresh18(fresh17(true2, true2, w0), true2, w0), true2, w0)
% 9.15/1.58 = { by axiom 4 (m__5078_4) }
% 9.15/1.58 fresh155(fresh18(true2, true2, w0), true2, w0)
% 9.15/1.58 = { by axiom 3 (m__4998_1) }
% 9.15/1.58 fresh155(true2, true2, w0)
% 9.15/1.58 = { by axiom 2 (m__3435_3) }
% 9.15/1.58 true2
% 9.15/1.58 % SZS output end Proof
% 9.15/1.58
% 9.15/1.58 RESULT: Theorem (the conjecture is true).
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