TSTP Solution File: SWV488+2 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV488+2 : 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 : n014.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 23:04:40 EDT 2023
% Result : Theorem 0.21s 0.49s
% Output : Proof 0.21s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SWV488+2 : 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
% 0.14/0.34 % Computer : n014.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Tue Aug 29 07:40:00 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.21/0.49 Command-line arguments: --no-flatten-goal
% 0.21/0.49
% 0.21/0.49 % SZS status Theorem
% 0.21/0.49
% 0.21/0.49 % SZS output start Proof
% 0.21/0.49 Take the following subset of the input axioms:
% 0.21/0.49 fof(int_leq, axiom, ![I, J]: (int_leq(I, J) <=> (int_less(I, J) | I=J))).
% 0.21/0.49 fof(lti, conjecture, ![I2, J2]: ((int_leq(int_one, I2) & (int_leq(I2, J2) & int_leq(J2, n))) => (I2=J2 => a(I2, J2)!=real_zero))).
% 0.21/0.49 fof(qil, hypothesis, ![I2, J2]: ((int_leq(int_one, I2) & (int_leq(I2, n) & (int_leq(int_one, J2) & int_leq(J2, n)))) => (![C]: ((int_less(int_zero, C) & I2=plus(J2, C)) => ![K]: ((int_leq(int_one, K) & int_leq(K, J2)) => a(plus(K, C), K)=lu(plus(K, C), K))) & (![K2]: ((int_leq(int_one, K2) & int_leq(K2, J2)) => a(K2, K2)=real_one) & ![C2]: ((int_less(int_zero, C2) & J2=plus(I2, C2)) => ![K2]: ((int_leq(int_one, K2) & int_leq(K2, I2)) => a(K2, plus(K2, C2))=real_zero)))))).
% 0.21/0.49 fof(real_constants, axiom, real_zero!=real_one).
% 0.21/0.49
% 0.21/0.49 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.50 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.50 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.50 fresh(y, y, x1...xn) = u
% 0.21/0.50 C => fresh(s, t, x1...xn) = v
% 0.21/0.50 where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.50 variables of u and v.
% 0.21/0.50 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.50 input problem has no model of domain size 1).
% 0.21/0.50
% 0.21/0.50 The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.50
% 0.21/0.50 Axiom 1 (lti_1): i = j.
% 0.21/0.50 Axiom 2 (lti): a(i, j) = real_zero.
% 0.21/0.50 Axiom 3 (int_leq): int_leq(X, X) = true2.
% 0.21/0.50 Axiom 4 (lti_3): int_leq(j, n) = true2.
% 0.21/0.50 Axiom 5 (lti_2): int_leq(int_one, i) = true2.
% 0.21/0.50 Axiom 6 (qil_2): fresh33(X, X, Y) = real_one.
% 0.21/0.50 Axiom 7 (qil_2): fresh31(X, X, Y, Z) = a(Z, Z).
% 0.21/0.50 Axiom 8 (qil_2): fresh32(X, X, Y, Z, W) = fresh33(int_leq(Y, n), true2, W).
% 0.21/0.50 Axiom 9 (qil_2): fresh30(X, X, Y, Z, W) = fresh31(int_leq(Z, n), true2, Y, W).
% 0.21/0.50 Axiom 10 (qil_2): fresh29(X, X, Y, Z, W) = fresh32(int_leq(W, Z), true2, Y, Z, W).
% 0.21/0.50 Axiom 11 (qil_2): fresh28(X, X, Y, Z, W) = fresh30(int_leq(int_one, Y), true2, Y, Z, W).
% 0.21/0.50 Axiom 12 (qil_2): fresh28(int_leq(int_one, X), true2, Y, Z, X) = fresh29(int_leq(int_one, Z), true2, Y, Z, X).
% 0.21/0.50
% 0.21/0.50 Lemma 13: int_leq(int_one, j) = true2.
% 0.21/0.50 Proof:
% 0.21/0.50 int_leq(int_one, j)
% 0.21/0.50 = { by axiom 1 (lti_1) R->L }
% 0.21/0.50 int_leq(int_one, i)
% 0.21/0.50 = { by axiom 5 (lti_2) }
% 0.21/0.50 true2
% 0.21/0.50
% 0.21/0.50 Goal 1 (real_constants): real_zero = real_one.
% 0.21/0.50 Proof:
% 0.21/0.50 real_zero
% 0.21/0.50 = { by axiom 2 (lti) R->L }
% 0.21/0.50 a(i, j)
% 0.21/0.50 = { by axiom 1 (lti_1) }
% 0.21/0.50 a(j, j)
% 0.21/0.50 = { by axiom 7 (qil_2) R->L }
% 0.21/0.50 fresh31(true2, true2, j, j)
% 0.21/0.50 = { by axiom 4 (lti_3) R->L }
% 0.21/0.50 fresh31(int_leq(j, n), true2, j, j)
% 0.21/0.50 = { by axiom 9 (qil_2) R->L }
% 0.21/0.50 fresh30(true2, true2, j, j, j)
% 0.21/0.50 = { by lemma 13 R->L }
% 0.21/0.50 fresh30(int_leq(int_one, j), true2, j, j, j)
% 0.21/0.50 = { by axiom 11 (qil_2) R->L }
% 0.21/0.50 fresh28(true2, true2, j, j, j)
% 0.21/0.50 = { by lemma 13 R->L }
% 0.21/0.50 fresh28(int_leq(int_one, j), true2, j, j, j)
% 0.21/0.50 = { by axiom 12 (qil_2) }
% 0.21/0.50 fresh29(int_leq(int_one, j), true2, j, j, j)
% 0.21/0.50 = { by lemma 13 }
% 0.21/0.50 fresh29(true2, true2, j, j, j)
% 0.21/0.50 = { by axiom 10 (qil_2) }
% 0.21/0.50 fresh32(int_leq(j, j), true2, j, j, j)
% 0.21/0.50 = { by axiom 3 (int_leq) }
% 0.21/0.50 fresh32(true2, true2, j, j, j)
% 0.21/0.50 = { by axiom 8 (qil_2) }
% 0.21/0.50 fresh33(int_leq(j, n), true2, j)
% 0.21/0.50 = { by axiom 4 (lti_3) }
% 0.21/0.50 fresh33(true2, true2, j)
% 0.21/0.50 = { by axiom 6 (qil_2) }
% 0.21/0.50 real_one
% 0.21/0.50 % SZS output end Proof
% 0.21/0.50
% 0.21/0.50 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------