%------------------------------------------------------------------------------
% File     : Twee---2.4.2
% Problem  : KLE150+1 : TPTP v8.1.2. Released v4.0.0.
% 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 : n024.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 05:36:04 EDT 2023

% Result   : Theorem 0.21s 0.40s
% Output   : Proof 0.21s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.12  % Problem  : KLE150+1 : TPTP v8.1.2. Released v4.0.0.
% 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 : n024.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
% 0.14/0.35  % WCLimit  : 300
% 0.14/0.35  % DateTime : Tue Aug 29 10:56:38 EDT 2023
% 0.14/0.35  % CPUTime  : 
% 0.21/0.40  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.21/0.40  
% 0.21/0.40  % SZS status Theorem
% 0.21/0.40  
% 0.21/0.40  % SZS output start Proof
% 0.21/0.40  Take the following subset of the input axioms:
% 0.21/0.40    fof(additive_commutativity, axiom, ![A, B]: addition(A, B)=addition(B, A)).
% 0.21/0.40    fof(goals, conjecture, ![X0]: strong_iteration(multiplication(X0, zero))=addition(one, multiplication(X0, zero))).
% 0.21/0.40    fof(infty_unfold1, axiom, ![A2]: strong_iteration(A2)=addition(multiplication(A2, strong_iteration(A2)), one)).
% 0.21/0.40    fof(left_annihilation, axiom, ![A2]: multiplication(zero, A2)=zero).
% 0.21/0.40    fof(multiplicative_associativity, axiom, ![C, A2, B2]: multiplication(A2, multiplication(B2, C))=multiplication(multiplication(A2, B2), C)).
% 0.21/0.40  
% 0.21/0.40  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.21/0.40  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.21/0.40  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.21/0.40    fresh(y, y, x1...xn) = u
% 0.21/0.40    C => fresh(s, t, x1...xn) = v
% 0.21/0.40  where fresh is a fresh function symbol and x1..xn are the free
% 0.21/0.40  variables of u and v.
% 0.21/0.40  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.21/0.40  input problem has no model of domain size 1).
% 0.21/0.40  
% 0.21/0.41  The encoding turns the above axioms into the following unit equations and goals:
% 0.21/0.41  
% 0.21/0.41  Axiom 1 (left_annihilation): multiplication(zero, X) = zero.
% 0.21/0.41  Axiom 2 (additive_commutativity): addition(X, Y) = addition(Y, X).
% 0.21/0.41  Axiom 3 (multiplicative_associativity): multiplication(X, multiplication(Y, Z)) = multiplication(multiplication(X, Y), Z).
% 0.21/0.41  Axiom 4 (infty_unfold1): strong_iteration(X) = addition(multiplication(X, strong_iteration(X)), one).
% 0.21/0.41  
% 0.21/0.41  Goal 1 (goals): strong_iteration(multiplication(x0, zero)) = addition(one, multiplication(x0, zero)).
% 0.21/0.41  Proof:
% 0.21/0.41    strong_iteration(multiplication(x0, zero))
% 0.21/0.41  = { by axiom 4 (infty_unfold1) }
% 0.21/0.41    addition(multiplication(multiplication(x0, zero), strong_iteration(multiplication(x0, zero))), one)
% 0.21/0.41  = { by axiom 2 (additive_commutativity) }
% 0.21/0.41    addition(one, multiplication(multiplication(x0, zero), strong_iteration(multiplication(x0, zero))))
% 0.21/0.41  = { by axiom 3 (multiplicative_associativity) R->L }
% 0.21/0.41    addition(one, multiplication(x0, multiplication(zero, strong_iteration(multiplication(x0, zero)))))
% 0.21/0.41  = { by axiom 1 (left_annihilation) }
% 0.21/0.41    addition(one, multiplication(x0, zero))
% 0.21/0.41  % SZS output end Proof
% 0.21/0.41  
% 0.21/0.41  RESULT: Theorem (the conjecture is true).
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