TSTP Solution File: SWV215+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV215+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 : 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 23:03:01 EDT 2023
% Result : Theorem 0.20s 0.59s
% Output : Proof 0.20s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.12 % Problem : SWV215+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.08/0.13 % 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 : n018.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
% 0.13/0.35 % CPULimit : 300
% 0.13/0.35 % WCLimit : 300
% 0.13/0.35 % DateTime : Tue Aug 29 08:59:32 EDT 2023
% 0.13/0.35 % CPUTime :
% 0.20/0.59 Command-line arguments: --no-flatten-goal
% 0.20/0.59
% 0.20/0.59 % SZS status Theorem
% 0.20/0.59
% 0.20/0.61 % SZS output start Proof
% 0.20/0.61 Take the following subset of the input axioms:
% 0.20/0.64 fof(quaternion_ds1_symm_0161, conjecture, ![A, B]: ((leq(n0, A) & (leq(n0, B) & (leq(A, n5) & leq(B, n5)))) => ((~(n0=A & n3=B) & (~(n0=A & n4=B) & (~(n0=A & n5=B) & (~(n0=B & n1=A) & (~(n0=B & n2=A) & (~(n0=B & n3=A) & (~(n0=B & n4=A) & (~(n0=B & n5=A) & (~(n1=A & n1=B) & (~(n1=A & n2=B) & (~(n1=A & n3=B) & (~(n1=A & n4=B) & (~(n1=A & n5=B) & (~(n1=B & n2=A) & (~(n1=B & n3=A) & (~(n1=B & n4=A) & (~(n1=B & n5=A) & (~(n2=A & n2=B) & (~(n2=A & n3=B) & (~(n2=A & n4=B) & (~(n2=A & n5=B) & (~(n2=B & n3=A) & (~(n2=B & n4=A) & (~(n2=B & n5=A) & (~(n3=A & n3=B) & (~(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) & (n0=A & (n2=A & (n2=B & n4=B)))))))))))))))))))))))))))))))))))) => n0=times(divide(n1, n400), a_select2(sigma, n2))))).
% 0.20/0.64
% 0.20/0.64 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.64 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.64 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.64 fresh(y, y, x1...xn) = u
% 0.20/0.64 C => fresh(s, t, x1...xn) = v
% 0.20/0.64 where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.64 variables of u and v.
% 0.20/0.64 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.64 input problem has no model of domain size 1).
% 0.20/0.64
% 0.20/0.64 The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.64
% 0.20/0.64 Axiom 1 (quaternion_ds1_symm_0161_3): n4 = b.
% 0.20/0.64 Axiom 2 (quaternion_ds1_symm_0161_1): n2 = b.
% 0.20/0.64 Axiom 3 (quaternion_ds1_symm_0161_2): n2 = a.
% 0.20/0.64 Axiom 4 (quaternion_ds1_symm_0161): n0 = a.
% 0.20/0.64
% 0.20/0.64 Lemma 5: n2 = n4.
% 0.20/0.64 Proof:
% 0.20/0.64 n2
% 0.20/0.64 = { by axiom 2 (quaternion_ds1_symm_0161_1) }
% 0.20/0.64 b
% 0.20/0.64 = { by axiom 1 (quaternion_ds1_symm_0161_3) R->L }
% 0.20/0.64 n4
% 0.20/0.64
% 0.20/0.64 Lemma 6: a = n4.
% 0.20/0.64 Proof:
% 0.20/0.64 a
% 0.20/0.64 = { by axiom 3 (quaternion_ds1_symm_0161_2) R->L }
% 0.20/0.64 n2
% 0.20/0.64 = { by lemma 5 }
% 0.20/0.64 n4
% 0.20/0.64
% 0.20/0.64 Lemma 7: n0 = n4.
% 0.20/0.64 Proof:
% 0.20/0.64 n0
% 0.20/0.64 = { by axiom 4 (quaternion_ds1_symm_0161) }
% 0.20/0.64 a
% 0.20/0.64 = { by lemma 6 }
% 0.20/0.64 n4
% 0.20/0.64
% 0.20/0.64 Goal 1 (quaternion_ds1_symm_0161_38): tuple2(n4, n4) = tuple2(b, a).
% 0.20/0.64 Proof:
% 0.20/0.64 tuple2(n4, n4)
% 0.20/0.64 = { by lemma 6 R->L }
% 0.20/0.64 tuple2(n4, a)
% 0.20/0.64 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.64 tuple2(b, a)
% 0.20/0.64
% 0.20/0.64 Goal 2 (quaternion_ds1_symm_0161_31): tuple2(n2, n4) = tuple2(a, b).
% 0.20/0.64 Proof:
% 0.20/0.64 tuple2(n2, n4)
% 0.20/0.65 = { by lemma 5 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(a, n4)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(a, b)
% 0.20/0.65
% 0.20/0.65 Goal 3 (quaternion_ds1_symm_0161_28): tuple2(n2, n4) = tuple2(b, a).
% 0.20/0.65 Proof:
% 0.20/0.65 tuple2(n2, n4)
% 0.20/0.65 = { by lemma 5 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(n4, a)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(b, a)
% 0.20/0.65
% 0.20/0.65 Goal 4 (quaternion_ds1_symm_0161_26): tuple2(n2, n2) = tuple2(b, a).
% 0.20/0.65 Proof:
% 0.20/0.65 tuple2(n2, n2)
% 0.20/0.65 = { by lemma 5 }
% 0.20/0.65 tuple2(n4, n2)
% 0.20/0.65 = { by lemma 5 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(n4, a)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(b, a)
% 0.20/0.65
% 0.20/0.65 Goal 5 (quaternion_ds1_symm_0161_15): tuple2(n0, n4) = tuple2(a, b).
% 0.20/0.65 Proof:
% 0.20/0.65 tuple2(n0, n4)
% 0.20/0.65 = { by lemma 7 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(a, n4)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(a, b)
% 0.20/0.65
% 0.20/0.65 Goal 6 (quaternion_ds1_symm_0161_12): tuple2(n0, n4) = tuple2(b, a).
% 0.20/0.65 Proof:
% 0.20/0.65 tuple2(n0, n4)
% 0.20/0.65 = { by lemma 7 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(n4, a)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(b, a)
% 0.20/0.65
% 0.20/0.65 Goal 7 (quaternion_ds1_symm_0161_10): tuple2(n0, n2) = tuple2(b, a).
% 0.20/0.65 Proof:
% 0.20/0.65 tuple2(n0, n2)
% 0.20/0.65 = { by lemma 5 }
% 0.20/0.65 tuple2(n0, n4)
% 0.20/0.65 = { by lemma 7 }
% 0.20/0.65 tuple2(n4, n4)
% 0.20/0.65 = { by lemma 6 R->L }
% 0.20/0.65 tuple2(n4, a)
% 0.20/0.65 = { by axiom 1 (quaternion_ds1_symm_0161_3) }
% 0.20/0.65 tuple2(b, a)
% 0.20/0.65 % SZS output end Proof
% 0.20/0.65
% 0.20/0.65 RESULT: Theorem (the conjecture is true).
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