TSTP Solution File: SWV228+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV228+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 : n028.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 23:03:05 EDT 2023
% Result : Theorem 0.19s 0.59s
% Output : Proof 0.19s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.12/0.12 % Problem : SWV228+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.12/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.33 % Computer : n028.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 10:41:25 EDT 2023
% 0.13/0.33 % CPUTime :
% 0.19/0.59 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.19/0.59
% 0.19/0.59 % SZS status Theorem
% 0.19/0.59
% 0.19/0.61 % SZS output start Proof
% 0.19/0.61 Take the following subset of the input axioms:
% 0.19/0.61 fof(quaternion_ds1_symm_0681, conjecture, (![B, A2]: ((leq(n0, A2) & (leq(n0, B) & (leq(A2, n5) & leq(B, n5)))) => a_select3(q_ds1_filter, A2, B)=a_select3(q_ds1_filter, B, A2)) & ![C, D]: ((leq(n0, C) & (leq(n0, D) & (leq(C, n2) & leq(D, n2)))) => a_select3(r_ds1_filter, C, D)=a_select3(r_ds1_filter, D, C))) => ![E, F]: ((leq(n0, E) & (leq(n0, F) & (leq(E, n5) & leq(F, n5)))) => ((~(n0=E & n4=F) & (~(n0=E & n5=F) & (~(n0=F & n4=E) & (~(n0=F & n5=E) & (~(n1=E & n4=F) & (~(n1=E & n5=F) & (~(n1=F & n3=E) & (~(n1=F & n4=E) & (~(n1=F & n5=E) & (~(n2=E & n4=F) & (~(n2=E & n5=F) & (~(n2=F & n3=E) & (~(n2=F & n4=E) & (~(n2=F & n5=E) & (~(n3=E & n3=F) & (~(n3=E & n4=F) & (~(n3=E & n5=F) & (~(n3=F & n4=E) & (~(n3=F & n5=E) & (~(n4=E & n4=F) & (~(n4=E & n5=F) & (~(n4=F & n5=E) & (~(n5=E & n5=F) & (n0=F & (n3=E & n3=F))))))))))))))))))))))))) => n0=a_select2(xinit_noise, n3)))).
% 0.19/0.61
% 0.19/0.61 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.61 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.61 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.61 fresh(y, y, x1...xn) = u
% 0.19/0.61 C => fresh(s, t, x1...xn) = v
% 0.19/0.61 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.61 variables of u and v.
% 0.19/0.61 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.61 input problem has no model of domain size 1).
% 0.19/0.61
% 0.19/0.61 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.61
% 0.19/0.61 Axiom 1 (quaternion_ds1_symm_0681): n0 = f.
% 0.19/0.61 Axiom 2 (quaternion_ds1_symm_0681_1): n3 = f.
% 0.19/0.61 Axiom 3 (quaternion_ds1_symm_0681_2): n3 = e.
% 0.19/0.61
% 0.19/0.61 Lemma 4: n3 = n0.
% 0.19/0.61 Proof:
% 0.19/0.61 n3
% 0.19/0.61 = { by axiom 2 (quaternion_ds1_symm_0681_1) }
% 0.19/0.61 f
% 0.19/0.61 = { by axiom 1 (quaternion_ds1_symm_0681) R->L }
% 0.19/0.61 n0
% 0.19/0.61
% 0.19/0.61 Goal 1 (quaternion_ds1_symm_0681_22): tuple2(n3, n3) = tuple2(f, e).
% 0.19/0.61 Proof:
% 0.19/0.61 tuple2(n3, n3)
% 0.19/0.61 = { by lemma 4 }
% 0.19/0.61 tuple2(n3, n0)
% 0.19/0.61 = { by lemma 4 }
% 0.19/0.61 tuple2(n0, n0)
% 0.19/0.61 = { by lemma 4 R->L }
% 0.19/0.61 tuple2(n0, n3)
% 0.19/0.61 = { by axiom 3 (quaternion_ds1_symm_0681_2) }
% 0.19/0.61 tuple2(n0, e)
% 0.19/0.61 = { by axiom 1 (quaternion_ds1_symm_0681) }
% 0.19/0.61 tuple2(f, e)
% 0.19/0.61 % SZS output end Proof
% 0.19/0.61
% 0.19/0.61 RESULT: Theorem (the conjecture is true).
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