TSTP Solution File: SWV220+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV220+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 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 : n007.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 23:03:03 EDT 2023
% Result : Theorem 1.53s 0.57s
% Output : Proof 1.53s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.06/0.11 % Problem : SWV220+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.06/0.12 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.33 % Computer : n007.cluster.edu
% 0.13/0.33 % Model : x86_64 x86_64
% 0.13/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.33 % Memory : 8042.1875MB
% 0.13/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.33 % CPULimit : 300
% 0.13/0.33 % WCLimit : 300
% 0.13/0.33 % DateTime : Tue Aug 29 04:29:28 EDT 2023
% 0.13/0.33 % CPUTime :
% 1.53/0.57 Command-line arguments: --lhs-weight 9 --flip-ordering --complete-subsets --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 1.53/0.57
% 1.53/0.57 % SZS status Theorem
% 1.53/0.57
% 1.53/0.57 % SZS output start Proof
% 1.53/0.57 Take the following subset of the input axioms:
% 1.53/0.59 fof(quaternion_ds1_symm_0361, conjecture, ![A, B]: ((leq(n0, A) & (leq(n0, B) & (leq(A, n5) & leq(B, n5)))) => ((~(n0=B & n4=A) & (~(n0=B & n5=A) & (~(n1=B & n4=A) & (~(n1=B & n5=A) & (~(n2=A & n5=B) & (~(n2=B & n4=A) & (~(n2=B & n5=A) & (~(n3=A & n4=B) & (~(n3=A & n5=B) & (~(n3=B & n4=A) & (~(n3=B & n5=A) & (~(n4=A & n4=B) & (~(n4=A & n5=B) & (~(n4=B & n5=A) & (~(n5=A & n5=B) & (n1=A & (n3=A & (n3=B & n5=B)))))))))))))))))) => times(divide(n1, n400), a_select2(sigma, n3))=n0))).
% 1.53/0.59
% 1.53/0.59 Now clausify the problem and encode Horn clauses using encoding 3 of
% 1.53/0.59 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 1.53/0.59 We repeatedly replace C & s=t => u=v by the two clauses:
% 1.53/0.59 fresh(y, y, x1...xn) = u
% 1.53/0.59 C => fresh(s, t, x1...xn) = v
% 1.53/0.59 where fresh is a fresh function symbol and x1..xn are the free
% 1.53/0.59 variables of u and v.
% 1.53/0.59 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 1.53/0.59 input problem has no model of domain size 1).
% 1.53/0.59
% 1.53/0.59 The encoding turns the above axioms into the following unit equations and goals:
% 1.53/0.59
% 1.53/0.59 Axiom 1 (quaternion_ds1_symm_0361_3): n5 = b.
% 1.53/0.59 Axiom 2 (quaternion_ds1_symm_0361_1): n3 = b.
% 1.53/0.59 Axiom 3 (quaternion_ds1_symm_0361_2): n3 = a.
% 1.53/0.59 Axiom 4 (quaternion_ds1_symm_0361): n1 = a.
% 1.53/0.59
% 1.53/0.59 Lemma 5: n3 = n5.
% 1.53/0.59 Proof:
% 1.53/0.59 n3
% 1.53/0.59 = { by axiom 2 (quaternion_ds1_symm_0361_1) }
% 1.53/0.59 b
% 1.53/0.59 = { by axiom 1 (quaternion_ds1_symm_0361_3) R->L }
% 1.53/0.59 n5
% 1.53/0.59
% 1.53/0.59 Lemma 6: a = n5.
% 1.53/0.59 Proof:
% 1.53/0.59 a
% 1.53/0.59 = { by axiom 3 (quaternion_ds1_symm_0361_2) R->L }
% 1.53/0.59 n3
% 1.53/0.59 = { by lemma 5 }
% 1.53/0.59 n5
% 1.53/0.59
% 1.53/0.59 Goal 1 (quaternion_ds1_symm_0361_23): tuple2(n5, n5) = tuple2(b, a).
% 1.53/0.59 Proof:
% 1.53/0.59 tuple2(n5, n5)
% 1.53/0.59 = { by lemma 6 R->L }
% 1.53/0.59 tuple2(n5, a)
% 1.53/0.59 = { by axiom 1 (quaternion_ds1_symm_0361_3) }
% 1.53/0.59 tuple2(b, a)
% 1.53/0.59
% 1.53/0.59 Goal 2 (quaternion_ds1_symm_0361_19): tuple2(n3, n5) = tuple2(a, b).
% 1.53/0.59 Proof:
% 1.53/0.59 tuple2(n3, n5)
% 1.53/0.59 = { by lemma 5 }
% 1.53/0.59 tuple2(n5, n5)
% 1.53/0.59 = { by lemma 6 R->L }
% 1.53/0.59 tuple2(a, n5)
% 1.53/0.59 = { by axiom 1 (quaternion_ds1_symm_0361_3) }
% 1.53/0.59 tuple2(a, b)
% 1.53/0.59
% 1.53/0.59 Goal 3 (quaternion_ds1_symm_0361_17): tuple2(n3, n5) = tuple2(b, a).
% 1.53/0.59 Proof:
% 1.53/0.59 tuple2(n3, n5)
% 1.53/0.59 = { by lemma 5 }
% 1.53/0.59 tuple2(n5, n5)
% 1.53/0.59 = { by lemma 6 R->L }
% 1.53/0.59 tuple2(n5, a)
% 1.53/0.59 = { by axiom 1 (quaternion_ds1_symm_0361_3) }
% 1.53/0.59 tuple2(b, a)
% 1.53/0.59
% 1.53/0.59 Goal 4 (quaternion_ds1_symm_0361_12): tuple2(n1, n5) = tuple2(b, a).
% 1.53/0.59 Proof:
% 1.53/0.59 tuple2(n1, n5)
% 1.53/0.59 = { by axiom 4 (quaternion_ds1_symm_0361) }
% 1.53/0.59 tuple2(a, n5)
% 1.53/0.59 = { by lemma 6 }
% 1.53/0.59 tuple2(n5, n5)
% 1.53/0.59 = { by lemma 6 R->L }
% 1.53/0.59 tuple2(n5, a)
% 1.53/0.59 = { by axiom 1 (quaternion_ds1_symm_0361_3) }
% 1.53/0.59 tuple2(b, a)
% 1.53/0.59 % SZS output end Proof
% 1.53/0.59
% 1.53/0.59 RESULT: Theorem (the conjecture is true).
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