TSTP Solution File: NUN088+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : NUN088+1 : TPTP v8.1.2. Released v7.3.0.
% 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 : n018.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 12:51:56 EDT 2023
% Result : Theorem 0.20s 0.52s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.13/0.15 % Problem : NUN088+1 : TPTP v8.1.2. Released v7.3.0.
% 0.13/0.16 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.15/0.37 % Computer : n018.cluster.edu
% 0.15/0.37 % Model : x86_64 x86_64
% 0.15/0.37 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.15/0.37 % Memory : 8042.1875MB
% 0.15/0.37 % OS : Linux 3.10.0-693.el7.x86_64
% 0.15/0.37 % CPULimit : 300
% 0.15/0.37 % WCLimit : 300
% 0.15/0.37 % DateTime : Sun Aug 27 09:15:30 EDT 2023
% 0.15/0.37 % CPUTime :
% 0.20/0.52 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 0.20/0.52
% 0.20/0.52 % SZS status Theorem
% 0.20/0.52
% 0.20/0.53 % SZS output start Proof
% 0.20/0.53 Take the following subset of the input axioms:
% 0.20/0.53 fof(axiom_7a, axiom, ![X7, Y10]: (![Y20]: (~id(Y20, Y10) | ~r1(Y20)) | ~r2(X7, Y10))).
% 0.20/0.53 fof(zerounidone, conjecture, ![Y1]: (![Y2]: (~id(Y2, Y1) | ~r1(Y2)) | ![Y3]: (~r1(Y3) | ~r2(Y3, Y1)))).
% 0.20/0.53
% 0.20/0.53 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.53 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.53 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.53 fresh(y, y, x1...xn) = u
% 0.20/0.53 C => fresh(s, t, x1...xn) = v
% 0.20/0.53 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.53 variables of u and v.
% 0.20/0.53 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.53 input problem has no model of domain size 1).
% 0.20/0.53
% 0.20/0.53 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.53
% 0.20/0.53 Axiom 1 (zerounidone_1): r1(y2) = true2.
% 0.20/0.53 Axiom 2 (zerounidone): id(y2, y1) = true2.
% 0.20/0.53 Axiom 3 (zerounidone_3): r2(y3, y1) = true2.
% 0.20/0.53
% 0.20/0.53 Goal 1 (axiom_7a): tuple(id(X, Y), r1(X), r2(Z, Y)) = tuple(true2, true2, true2).
% 0.20/0.53 The goal is true when:
% 0.20/0.53 X = y2
% 0.20/0.53 Y = y1
% 0.20/0.53 Z = y3
% 0.20/0.53
% 0.20/0.53 Proof:
% 0.20/0.53 tuple(id(y2, y1), r1(y2), r2(y3, y1))
% 0.20/0.53 = { by axiom 2 (zerounidone) }
% 0.20/0.53 tuple(true2, r1(y2), r2(y3, y1))
% 0.20/0.53 = { by axiom 1 (zerounidone_1) }
% 0.20/0.53 tuple(true2, true2, r2(y3, y1))
% 0.20/0.53 = { by axiom 3 (zerounidone_3) }
% 0.20/0.53 tuple(true2, true2, true2)
% 0.20/0.53 % SZS output end Proof
% 0.20/0.53
% 0.20/0.53 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------