TSTP Solution File: SWV128+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWV128+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 : 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 : Thu Aug 31 23:02:41 EDT 2023
% Result : Theorem 0.21s 0.62s
% Output : Proof 0.21s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.11/0.13 % Problem : SWV128+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
% 0.14/0.35 % Computer : n011.cluster.edu
% 0.14/0.35 % Model : x86_64 x86_64
% 0.14/0.35 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.35 % Memory : 8042.1875MB
% 0.14/0.35 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.35 % CPULimit : 300
% 0.14/0.35 % WCLimit : 300
% 0.14/0.35 % DateTime : Tue Aug 29 03:33:56 EDT 2023
% 0.14/0.35 % CPUTime :
% 0.21/0.62 Command-line arguments: --ground-connectedness --complete-subsets
% 0.21/0.62
% 0.21/0.62 % SZS status Theorem
% 0.21/0.62
% 0.21/0.62 % SZS output start Proof
% 0.21/0.62 Take the following subset of the input axioms:
% 0.21/0.62 fof(thruster_inuse_0001, conjecture, (t_defuse=use & (tvar_defuse=use & (![B, A2]: ((leq(n0, A2) & (leq(n0, B) & (leq(A2, minus(m_measvars, n1)) & leq(B, minus(n_steps, n1))))) => a_select3(rho_defuse, A2, B)=use) & (![C, D]: ((leq(n0, C) & (leq(n0, D) & (leq(C, n7) & leq(D, minus(n_steps, n1))))) => a_select3(tr_defuse, C, D)=use) & (![E]: ((leq(n0, E) & leq(E, minus(n_statevars, n1))) => a_select2(xinit_defuse, E)=use) & (![F]: ((leq(n0, F) & leq(F, minus(n_statevars, n1))) => a_select2(xinit_mean_defuse, F)=use) & ![G, H]: ((leq(n0, G) & (leq(n0, H) & (leq(G, minus(m_measvars, n1)) & leq(H, minus(n_steps, n1))))) => a_select3(z_defuse, G, H)=use))))))) => true).
% 0.21/0.62 fof(ttrue, axiom, true).
% 0.21/0.62
% 0.21/0.62 Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.62 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.62 We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.62 fresh(y, y, x1...xn) = u
% 0.21/0.62 C => fresh(s, t, x1...xn) = v
% 0.21/0.62 where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.62 variables of u and v.
% 0.21/0.62 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.62 input problem has no model of domain size 1).
% 0.21/0.62
% 0.21/0.62 The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.62
% 0.21/0.62 Axiom 1 (ttrue): true = true3.
% 0.21/0.62
% 0.21/0.62 Goal 1 (thruster_inuse_0001_7): true = true3.
% 0.21/0.62 Proof:
% 0.21/0.62 true
% 0.21/0.62 = { by axiom 1 (ttrue) }
% 0.21/0.62 true3
% 0.21/0.62 % SZS output end Proof
% 0.21/0.62
% 0.21/0.62 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------