TSTP Solution File: ALG200+1 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : ALG200+1 : TPTP v8.1.2. Released v2.7.0.
% 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 : n011.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 : Wed Aug 30 16:42:28 EDT 2023
% Result : Theorem 0.19s 0.51s
% 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
% 0.10/0.12 % Problem : ALG200+1 : TPTP v8.1.2. Released v2.7.0.
% 0.10/0.13 % 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 : n011.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
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Mon Aug 28 03:23:55 EDT 2023
% 0.13/0.34 % CPUTime :
% 0.19/0.51 Command-line arguments: --set-join --lhs-weight 1 --no-flatten-goal --complete-subsets --goal-heuristic
% 0.19/0.51
% 0.19/0.51 % SZS status Theorem
% 0.19/0.51
% 0.19/0.56 % SZS output start Proof
% 0.19/0.56 Take the following subset of the input axioms:
% 0.19/0.56 fof(co1, conjecture, ~(op(e0, e0)=e0 & (op(e1, e1)=e1 & (op(e2, e2)=e2 & (op(e3, e3)=e3 & (op(e4, e4)=e4 & (op(e5, e5)=e5 & (op(e6, e6)=e6 & ((op(e0, e0)=e0 | (op(e1, e1)=e1 | (op(e2, e2)=e2 | (op(e3, e3)=e3 | (op(e4, e4)=e4 | (op(e5, e5)=e5 | op(e6, e6)=e6)))))) & (op(e0, e0)!=e0 | (op(e1, e1)!=e1 | (op(e2, e2)!=e2 | (op(e3, e3)!=e3 | (op(e4, e4)!=e4 | (op(e5, e5)!=e5 | op(e6, e6)!=e6))))))))))))))).
% 0.19/0.56
% 0.19/0.56 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.56 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.56 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.56 fresh(y, y, x1...xn) = u
% 0.19/0.56 C => fresh(s, t, x1...xn) = v
% 0.19/0.56 where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.56 variables of u and v.
% 0.19/0.56 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.56 input problem has no model of domain size 1).
% 0.19/0.56
% 0.19/0.56 The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.56
% 0.19/0.56 Axiom 1 (co1): op(e0, e0) = e0.
% 0.19/0.56 Axiom 2 (co1_3): op(e2, e2) = e2.
% 0.19/0.56 Axiom 3 (co1_5): op(e4, e4) = e4.
% 0.19/0.56 Axiom 4 (co1_6): op(e5, e5) = e5.
% 0.19/0.56 Axiom 5 (co1_7): op(e6, e6) = e6.
% 0.19/0.56 Axiom 6 (co1_2): op(e1, e1) = e1.
% 0.19/0.56 Axiom 7 (co1_4): op(e3, e3) = e3.
% 0.19/0.56
% 0.19/0.56 Goal 1 (co1_8): tuple(op(e0, e0), op(e1, e1), op(e2, e2), op(e3, e3), op(e4, e4), op(e5, e5), op(e6, e6)) = tuple(e0, e1, e2, e3, e4, e5, e6).
% 0.19/0.56 Proof:
% 0.19/0.56 tuple(op(e0, e0), op(e1, e1), op(e2, e2), op(e3, e3), op(e4, e4), op(e5, e5), op(e6, e6))
% 0.19/0.56 = { by axiom 1 (co1) }
% 0.19/0.56 tuple(e0, op(e1, e1), op(e2, e2), op(e3, e3), op(e4, e4), op(e5, e5), op(e6, e6))
% 0.19/0.56 = { by axiom 2 (co1_3) }
% 0.19/0.56 tuple(e0, op(e1, e1), e2, op(e3, e3), op(e4, e4), op(e5, e5), op(e6, e6))
% 0.19/0.56 = { by axiom 3 (co1_5) }
% 0.19/0.56 tuple(e0, op(e1, e1), e2, op(e3, e3), e4, op(e5, e5), op(e6, e6))
% 0.19/0.56 = { by axiom 4 (co1_6) }
% 0.19/0.56 tuple(e0, op(e1, e1), e2, op(e3, e3), e4, e5, op(e6, e6))
% 0.19/0.56 = { by axiom 5 (co1_7) }
% 0.19/0.56 tuple(e0, op(e1, e1), e2, op(e3, e3), e4, e5, e6)
% 0.19/0.56 = { by axiom 6 (co1_2) }
% 0.19/0.56 tuple(e0, e1, e2, op(e3, e3), e4, e5, e6)
% 0.19/0.56 = { by axiom 7 (co1_4) }
% 0.19/0.56 tuple(e0, e1, e2, e3, e4, e5, e6)
% 0.19/0.56 % SZS output end Proof
% 0.19/0.56
% 0.19/0.56 RESULT: Theorem (the conjecture is true).
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