TSTP Solution File: NUM552+3 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUM552+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 : n004.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 11:56:56 EDT 2023
% Result : Theorem 2.49s 0.86s
% Output : Proof 2.83s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : NUM552+3 : 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
% 0.11/0.32 % Computer : n004.cluster.edu
% 0.11/0.32 % Model : x86_64 x86_64
% 0.11/0.32 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.11/0.32 % Memory : 8042.1875MB
% 0.11/0.32 % OS : Linux 3.10.0-693.el7.x86_64
% 0.11/0.32 % CPULimit : 300
% 0.11/0.32 % WCLimit : 300
% 0.11/0.32 % DateTime : Fri Aug 25 08:54:53 EDT 2023
% 0.11/0.32 % CPUTime :
% 2.49/0.86 Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 2.49/0.86
% 2.49/0.86 % SZS status Theorem
% 2.49/0.87
% 2.49/0.87 % SZS output start Proof
% 2.49/0.87 Take the following subset of the input axioms:
% 2.49/0.88 fof(m__, conjecture, aElementOf0(xx, xQ) => aElementOf0(xx, xT)).
% 2.49/0.88 fof(m__2227, hypothesis, aSet0(slbdtsldtrb0(xS, xk)) & (![W0]: ((aElementOf0(W0, slbdtsldtrb0(xS, xk)) => (aSet0(W0) & (![W1]: (aElementOf0(W1, W0) => aElementOf0(W1, xS)) & (aSubsetOf0(W0, xS) & sbrdtbr0(W0)=xk)))) & ((((aSet0(W0) & ![W1_2]: (aElementOf0(W1_2, W0) => aElementOf0(W1_2, xS))) | aSubsetOf0(W0, xS)) & sbrdtbr0(W0)=xk) => aElementOf0(W0, slbdtsldtrb0(xS, xk)))) & (aSet0(slbdtsldtrb0(xT, xk)) & (![W0_2]: ((aElementOf0(W0_2, slbdtsldtrb0(xT, xk)) => (aSet0(W0_2) & (![W1_2]: (aElementOf0(W1_2, W0_2) => aElementOf0(W1_2, xT)) & (aSubsetOf0(W0_2, xT) & sbrdtbr0(W0_2)=xk)))) & ((((aSet0(W0_2) & ![W1_2]: (aElementOf0(W1_2, W0_2) => aElementOf0(W1_2, xT))) | aSubsetOf0(W0_2, xT)) & sbrdtbr0(W0_2)=xk) => aElementOf0(W0_2, slbdtsldtrb0(xT, xk)))) & (![W0_2]: (aElementOf0(W0_2, slbdtsldtrb0(xS, xk)) => aElementOf0(W0_2, slbdtsldtrb0(xT, xk))) & (aSubsetOf0(slbdtsldtrb0(xS, xk), slbdtsldtrb0(xT, xk)) & ~(![W0_2]: ((aElementOf0(W0_2, slbdtsldtrb0(xS, xk)) => (aSet0(W0_2) & (![W1_2]: (aElementOf0(W1_2, W0_2) => aElementOf0(W1_2, xS)) & (aSubsetOf0(W0_2, xS) & sbrdtbr0(W0_2)=xk)))) & ((((aSet0(W0_2) & ![W1_2]: (aElementOf0(W1_2, W0_2) => aElementOf0(W1_2, xS))) | aSubsetOf0(W0_2, xS)) & sbrdtbr0(W0_2)=xk) => aElementOf0(W0_2, slbdtsldtrb0(xS, xk)))) => (~?[W0_2]: aElementOf0(W0_2, slbdtsldtrb0(xS, xk)) | slbdtsldtrb0(xS, xk)=slcrc0)))))))).
% 2.49/0.88 fof(m__2270, hypothesis, aSet0(xQ) & (![W0_2]: (aElementOf0(W0_2, xQ) => aElementOf0(W0_2, xS)) & (aSubsetOf0(xQ, xS) & (sbrdtbr0(xQ)=xk & aElementOf0(xQ, slbdtsldtrb0(xS, xk)))))).
% 2.83/0.89
% 2.83/0.89 Now clausify the problem and encode Horn clauses using encoding 3 of
% 2.83/0.89 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 2.83/0.89 We repeatedly replace C & s=t => u=v by the two clauses:
% 2.83/0.89 fresh(y, y, x1...xn) = u
% 2.83/0.89 C => fresh(s, t, x1...xn) = v
% 2.83/0.89 where fresh is a fresh function symbol and x1..xn are the free
% 2.83/0.89 variables of u and v.
% 2.83/0.89 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 2.83/0.89 input problem has no model of domain size 1).
% 2.83/0.89
% 2.83/0.89 The encoding turns the above axioms into the following unit equations and goals:
% 2.83/0.89
% 2.83/0.89 Axiom 1 (m__): aElementOf0(xx, xQ) = true2.
% 2.83/0.89 Axiom 2 (m__2227_21): fresh17(X, X, Y) = true2.
% 2.83/0.89 Axiom 3 (m__2227_22): fresh16(X, X, Y) = true2.
% 2.83/0.89 Axiom 4 (m__2270_2): aElementOf0(xQ, slbdtsldtrb0(xS, xk)) = true2.
% 2.83/0.89 Axiom 5 (m__2227_21): fresh18(X, X, Y, Z) = aElementOf0(Z, xT).
% 2.83/0.89 Axiom 6 (m__2227_21): fresh18(aElementOf0(X, Y), true2, Y, X) = fresh17(aElementOf0(Y, slbdtsldtrb0(xT, xk)), true2, X).
% 2.83/0.89 Axiom 7 (m__2227_22): fresh16(aElementOf0(X, slbdtsldtrb0(xS, xk)), true2, X) = aElementOf0(X, slbdtsldtrb0(xT, xk)).
% 2.83/0.89
% 2.83/0.89 Goal 1 (m___1): aElementOf0(xx, xT) = true2.
% 2.83/0.89 Proof:
% 2.83/0.89 aElementOf0(xx, xT)
% 2.83/0.89 = { by axiom 5 (m__2227_21) R->L }
% 2.83/0.89 fresh18(true2, true2, xQ, xx)
% 2.83/0.89 = { by axiom 1 (m__) R->L }
% 2.83/0.89 fresh18(aElementOf0(xx, xQ), true2, xQ, xx)
% 2.83/0.89 = { by axiom 6 (m__2227_21) }
% 2.83/0.89 fresh17(aElementOf0(xQ, slbdtsldtrb0(xT, xk)), true2, xx)
% 2.83/0.89 = { by axiom 7 (m__2227_22) R->L }
% 2.83/0.89 fresh17(fresh16(aElementOf0(xQ, slbdtsldtrb0(xS, xk)), true2, xQ), true2, xx)
% 2.83/0.89 = { by axiom 4 (m__2270_2) }
% 2.83/0.89 fresh17(fresh16(true2, true2, xQ), true2, xx)
% 2.83/0.89 = { by axiom 3 (m__2227_22) }
% 2.83/0.89 fresh17(true2, true2, xx)
% 2.83/0.89 = { by axiom 2 (m__2227_21) }
% 2.83/0.89 true2
% 2.83/0.89 % SZS output end Proof
% 2.83/0.89
% 2.83/0.89 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------