%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : SWV121+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 : n026.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 23:02:40 EDT 2023

% Result   : Theorem 0.20s 0.60s
% Output   : Proof 0.20s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : SWV121+1 : TPTP v8.1.2. Bugfixed v3.3.0.
% 0.00/0.13  % 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 : n026.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 03:53:20 EDT 2023
% 0.13/0.34  % CPUTime  : 
% 0.20/0.60  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.20/0.60  
% 0.20/0.60  % SZS status Theorem
% 0.20/0.60  
% 0.20/0.60  % SZS output start Proof
% 0.20/0.60  Take the following subset of the input axioms:
% 0.20/0.60    fof(quaternion_ds1_symm_0014, conjecture, (![B, A2]: ((leq(n0, A2) & (leq(n0, B) & (leq(A2, minus(n6, n1)) & leq(B, minus(n6, n1))))) => a_select3(q_ds1_filter, A2, B)=a_select3(q_ds1_filter, B, A2)) & (![C, D]: ((leq(n0, C) & (leq(n0, D) & (leq(C, minus(n3, n1)) & leq(D, minus(n3, n1))))) => a_select3(r_ds1_filter, C, D)=a_select3(r_ds1_filter, D, C)) & ![E, F]: ((leq(n0, E) & (leq(n0, F) & (leq(E, minus(n6, n1)) & leq(F, minus(n6, n1))))) => a_select3(pminus_ds1_filter, E, F)=a_select3(pminus_ds1_filter, F, E)))) => true).
% 0.20/0.60    fof(ttrue, axiom, true).
% 0.20/0.60  
% 0.20/0.60  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.20/0.60  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.20/0.60  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.20/0.60    fresh(y, y, x1...xn) = u
% 0.20/0.60    C => fresh(s, t, x1...xn) = v
% 0.20/0.60  where fresh is a fresh function symbol and x1..xn are the free
% 0.20/0.60  variables of u and v.
% 0.20/0.60  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.20/0.60  input problem has no model of domain size 1).
% 0.20/0.60  
% 0.20/0.60  The encoding turns the above axioms into the following unit equations and goals:
% 0.20/0.60  
% 0.20/0.60  Axiom 1 (ttrue): true = true3.
% 0.20/0.60  
% 0.20/0.60  Goal 1 (quaternion_ds1_symm_0014_3): true = true3.
% 0.20/0.60  Proof:
% 0.20/0.60    true
% 0.20/0.61  = { by axiom 1 (ttrue) }
% 0.20/0.61    true3
% 0.20/0.61  % SZS output end Proof
% 0.20/0.61  
% 0.20/0.61  RESULT: Theorem (the conjecture is true).
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