TSTP Solution File: SWV232+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV232+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 : n006.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 23:03:06 EDT 2023
% Result : Theorem 0.18s 0.58s
% Output : Proof 0.18s
% Verified :
% SZS Type : -
% Comments :
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.12 % Problem : SWV232+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.11/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.33 % Computer : n006.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Tue Aug 29 09:52:06 EDT 2023
% 0.12/0.34 % CPUTime :
% 0.18/0.58 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.18/0.58
% 0.18/0.58 % SZS status Theorem
% 0.18/0.58
% 0.18/0.59 % SZS output start Proof
% 0.18/0.59 Take the following subset of the input axioms:
% 0.18/0.60 fof(quaternion_ds1_symm_0841, 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)))) => ((~(n0=C & n1=D) & (~(n0=C & n2=D) & (~(n0=D & n2=C) & (~(n1=C & n1=D) & (~(n1=C & n2=D) & (~(n1=D & n2=C) & (~(n2=C & n2=D) & (n0=D & (n1=C & (n1=D & n2=C)))))))))) => a_select2(rho, n1)=n0))).
% 0.18/0.60
% 0.18/0.60 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.18/0.60 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.18/0.60 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.18/0.60 fresh(y, y, x1...xn) = u
% 0.18/0.60 C => fresh(s, t, x1...xn) = v
% 0.18/0.60 where fresh is a fresh function symbol and x1..xn are the free
% 0.18/0.60 variables of u and v.
% 0.18/0.60 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.18/0.60 input problem has no model of domain size 1).
% 0.18/0.60
% 0.18/0.60 The encoding turns the above axioms into the following unit equations and goals:
% 0.18/0.60
% 0.18/0.60 Axiom 1 (quaternion_ds1_symm_0841): n0 = d.
% 0.18/0.60 Axiom 2 (quaternion_ds1_symm_0841_1): n1 = d.
% 0.18/0.60 Axiom 3 (quaternion_ds1_symm_0841_2): n1 = c.
% 0.18/0.60 Axiom 4 (quaternion_ds1_symm_0841_3): n2 = c.
% 0.18/0.60
% 0.18/0.60 Lemma 5: n1 = n0.
% 0.18/0.60 Proof:
% 0.18/0.60 n1
% 0.18/0.60 = { by axiom 2 (quaternion_ds1_symm_0841_1) }
% 0.18/0.60 d
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) R->L }
% 0.18/0.60 n0
% 0.18/0.60
% 0.18/0.60 Lemma 6: c = n0.
% 0.18/0.60 Proof:
% 0.18/0.60 c
% 0.18/0.60 = { by axiom 3 (quaternion_ds1_symm_0841_2) R->L }
% 0.18/0.60 n1
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 n0
% 0.18/0.60
% 0.18/0.60 Lemma 7: n2 = n0.
% 0.18/0.60 Proof:
% 0.18/0.60 n2
% 0.18/0.60 = { by axiom 4 (quaternion_ds1_symm_0841_3) }
% 0.18/0.60 c
% 0.18/0.60 = { by lemma 6 }
% 0.18/0.60 n0
% 0.18/0.60
% 0.18/0.60 Goal 1 (quaternion_ds1_symm_0841_15): tuple2(n2, n2) = tuple2(d, c).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n2, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(n0, c)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(d, c)
% 0.18/0.60
% 0.18/0.60 Goal 2 (quaternion_ds1_symm_0841_14): tuple2(n1, n2) = tuple2(c, d).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n1, n2)
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 tuple2(n0, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(c, n0)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(c, d)
% 0.18/0.60
% 0.18/0.60 Goal 3 (quaternion_ds1_symm_0841_13): tuple2(n1, n2) = tuple2(d, c).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n1, n2)
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 tuple2(n0, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(n0, c)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(d, c)
% 0.18/0.60
% 0.18/0.60 Goal 4 (quaternion_ds1_symm_0841_12): tuple2(n1, n1) = tuple2(d, c).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n1, n1)
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 tuple2(n0, n1)
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(n0, c)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(d, c)
% 0.18/0.60
% 0.18/0.60 Goal 5 (quaternion_ds1_symm_0841_11): tuple2(n0, n2) = tuple2(c, d).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n0, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(c, n0)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(c, d)
% 0.18/0.60
% 0.18/0.60 Goal 6 (quaternion_ds1_symm_0841_10): tuple2(n0, n1) = tuple2(c, d).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n0, n1)
% 0.18/0.60 = { by lemma 5 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(c, n0)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(c, d)
% 0.18/0.60
% 0.18/0.60 Goal 7 (quaternion_ds1_symm_0841_9): tuple2(n0, n2) = tuple2(d, c).
% 0.18/0.60 Proof:
% 0.18/0.60 tuple2(n0, n2)
% 0.18/0.60 = { by lemma 7 }
% 0.18/0.60 tuple2(n0, n0)
% 0.18/0.60 = { by lemma 6 R->L }
% 0.18/0.60 tuple2(n0, c)
% 0.18/0.60 = { by axiom 1 (quaternion_ds1_symm_0841) }
% 0.18/0.60 tuple2(d, c)
% 0.18/0.60 % SZS output end Proof
% 0.18/0.60
% 0.18/0.60 RESULT: Theorem (the conjecture is true).
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