TSTP Solution File: SWC291+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWC291+1 : TPTP v8.1.2. Released v2.4.0.
% 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 20:54:49 EDT 2023
% Result : Theorem 3.76s 0.85s
% Output : Proof 3.76s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12 % Problem : SWC291+1 : TPTP v8.1.2. Released v2.4.0.
% 0.07/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.34 % Computer : n028.cluster.edu
% 0.14/0.34 % Model : x86_64 x86_64
% 0.14/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.34 % Memory : 8042.1875MB
% 0.14/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.34 % CPULimit : 300
% 0.14/0.34 % WCLimit : 300
% 0.14/0.34 % DateTime : Mon Aug 28 18:50:39 EDT 2023
% 0.14/0.34 % CPUTime :
% 3.76/0.85 Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 3.76/0.85
% 3.76/0.85 % SZS status Theorem
% 3.76/0.85
% 3.76/0.85 % SZS output start Proof
% 3.76/0.85 Take the following subset of the input axioms:
% 3.76/0.85 fof(co1, conjecture, ![U]: (ssList(U) => ![V]: (ssList(V) => ![W]: (ssList(W) => ![X]: (ssList(X) => (V!=X | (U!=W | (~segmentP(X, W) | (~strictorderedP(W) | (?[Y]: (ssList(Y) & (neq(W, Y) & (segmentP(X, Y) & (segmentP(Y, W) & strictorderedP(Y))))) | strictorderedP(U))))))))))).
% 3.76/0.85
% 3.76/0.85 Now clausify the problem and encode Horn clauses using encoding 3 of
% 3.76/0.85 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 3.76/0.85 We repeatedly replace C & s=t => u=v by the two clauses:
% 3.76/0.85 fresh(y, y, x1...xn) = u
% 3.76/0.85 C => fresh(s, t, x1...xn) = v
% 3.76/0.85 where fresh is a fresh function symbol and x1..xn are the free
% 3.76/0.85 variables of u and v.
% 3.76/0.85 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 3.76/0.85 input problem has no model of domain size 1).
% 3.76/0.85
% 3.76/0.85 The encoding turns the above axioms into the following unit equations and goals:
% 3.76/0.85
% 3.76/0.85 Axiom 1 (co1): u = w.
% 3.76/0.85 Axiom 2 (co1_7): strictorderedP(w) = true2.
% 3.76/0.85
% 3.76/0.85 Goal 1 (co1_9): strictorderedP(u) = true2.
% 3.76/0.85 Proof:
% 3.76/0.85 strictorderedP(u)
% 3.76/0.85 = { by axiom 1 (co1) }
% 3.76/0.85 strictorderedP(w)
% 3.76/0.85 = { by axiom 2 (co1_7) }
% 3.76/0.85 true2
% 3.76/0.85 % SZS output end Proof
% 3.76/0.85
% 3.76/0.85 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------