%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWV106+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 : n022.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 23:02:37 EDT 2023

% Result   : Theorem 0.20s 0.57s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.12  % Problem  : SWV106+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.08/0.13  % 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 : n022.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 : Tue Aug 29 03:54:09 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.20/0.57  Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 0.20/0.57  
% 0.20/0.57  % SZS status Theorem
% 0.20/0.57  
% 0.20/0.57  % SZS output start Proof
% 0.20/0.57  Take the following subset of the input axioms:
% 0.20/0.57    fof(quaternion_ds1_inuse_0018, conjecture, (![A2]: ((leq(n0, A2) & leq(A2, minus(m_measvars, n1))) => a_select2(rho_defuse, A2)=use) & (![B]: ((leq(n0, B) & leq(B, minus(n_statevars, n1))) => a_select2(sigma_defuse, B)=use) & (![C, D]: ((leq(n0, C) & (leq(n0, D) & (leq(C, n2) & leq(D, minus(n_steps, n1))))) => a_select3(u_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]: ((leq(n0, G) & leq(G, minus(n_statevars, n1))) => a_select2(xinit_noise_defuse, G)=use) & ![I, H]: ((leq(n0, H) & (leq(n0, I) & (leq(H, minus(m_measvars, n1)) & leq(I, minus(n_steps, n1))))) => a_select3(z_defuse, H, I)=use))))))) => true).
% 0.20/0.57    fof(ttrue, axiom, true).
% 0.20/0.57  
% 0.20/0.57  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.57  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.57  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.57    fresh(y, y, x1...xn) = u
% 0.20/0.57    C => fresh(s, t, x1...xn) = v
% 0.20/0.57  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.57  variables of u and v.
% 0.20/0.57  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.57  input problem has no model of domain size 1).
% 0.20/0.57  
% 0.20/0.57  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.57  
% 0.20/0.57  Axiom 1 (ttrue): true = true3.
% 0.20/0.57  
% 0.20/0.57  Goal 1 (quaternion_ds1_inuse_0018_7): true = true3.
% 0.20/0.57  Proof:
% 0.20/0.57    true
% 0.20/0.57  = { by axiom 1 (ttrue) }
% 0.20/0.57    true3
% 0.20/0.57  % SZS output end Proof
% 0.20/0.57  
% 0.20/0.57  RESULT: Theorem (the conjecture is true).
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