TSTP Solution File: KLE139+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : KLE139+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 : n007.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 05:36:00 EDT 2023

% Result   : Theorem 1.75s 0.57s
% Output   : Proof 1.75s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.04/0.12  % Problem  : KLE139+1 : TPTP v8.1.2. Released v4.0.0.
% 0.04/0.13  % 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 : n007.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 : Tue Aug 29 11:04:13 EDT 2023
% 0.14/0.34  % CPUTime  : 
% 1.75/0.57  Command-line arguments: --flip-ordering --lhs-weight 1 --depth-weight 60 --distributivity-heuristic
% 1.75/0.57  
% 1.75/0.57  % SZS status Theorem
% 1.75/0.57  
% 1.75/0.58  % SZS output start Proof
% 1.75/0.58  Take the following subset of the input axioms:
% 1.75/0.58    fof(additive_associativity, axiom, ![A, B, C]: addition(A, addition(B, C))=addition(addition(A, B), C)).
% 1.75/0.58    fof(additive_commutativity, axiom, ![A2, B2]: addition(A2, B2)=addition(B2, A2)).
% 1.75/0.58    fof(distributivity2, axiom, ![A2, B2, C2]: multiplication(addition(A2, B2), C2)=addition(multiplication(A2, C2), multiplication(B2, C2))).
% 1.75/0.58    fof(goals, conjecture, ![X0]: strong_iteration(X0)=addition(multiplication(strong_iteration(X0), X0), one)).
% 1.75/0.58    fof(isolation, axiom, ![A2]: strong_iteration(A2)=addition(star(A2), multiplication(strong_iteration(A2), zero))).
% 1.75/0.58    fof(left_annihilation, axiom, ![A2]: multiplication(zero, A2)=zero).
% 1.75/0.58    fof(multiplicative_associativity, axiom, ![A2, B2, C2]: multiplication(A2, multiplication(B2, C2))=multiplication(multiplication(A2, B2), C2)).
% 1.75/0.58    fof(star_unfold2, axiom, ![A2]: addition(one, multiplication(star(A2), A2))=star(A2)).
% 1.75/0.58  
% 1.75/0.58  Now clausify the problem and encode Horn clauses using encoding 3 of
% 1.75/0.58  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 1.75/0.58  We repeatedly replace C & s=t => u=v by the two clauses:
% 1.75/0.58    fresh(y, y, x1...xn) = u
% 1.75/0.58    C => fresh(s, t, x1...xn) = v
% 1.75/0.58  where fresh is a fresh function symbol and x1..xn are the free
% 1.75/0.58  variables of u and v.
% 1.75/0.58  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 1.75/0.58  input problem has no model of domain size 1).
% 1.75/0.58  
% 1.75/0.58  The encoding turns the above axioms into the following unit equations and goals:
% 1.75/0.58  
% 1.75/0.58  Axiom 1 (left_annihilation): multiplication(zero, X) = zero.
% 1.75/0.58  Axiom 2 (additive_commutativity): addition(X, Y) = addition(Y, X).
% 1.75/0.58  Axiom 3 (multiplicative_associativity): multiplication(X, multiplication(Y, Z)) = multiplication(multiplication(X, Y), Z).
% 1.75/0.58  Axiom 4 (additive_associativity): addition(X, addition(Y, Z)) = addition(addition(X, Y), Z).
% 1.75/0.58  Axiom 5 (star_unfold2): addition(one, multiplication(star(X), X)) = star(X).
% 1.75/0.58  Axiom 6 (isolation): strong_iteration(X) = addition(star(X), multiplication(strong_iteration(X), zero)).
% 1.75/0.58  Axiom 7 (distributivity2): multiplication(addition(X, Y), Z) = addition(multiplication(X, Z), multiplication(Y, Z)).
% 1.75/0.58  
% 1.75/0.58  Lemma 8: addition(multiplication(strong_iteration(X), zero), star(X)) = strong_iteration(X).
% 1.75/0.58  Proof:
% 1.75/0.58    addition(multiplication(strong_iteration(X), zero), star(X))
% 1.75/0.58  = { by axiom 2 (additive_commutativity) R->L }
% 1.75/0.58    addition(star(X), multiplication(strong_iteration(X), zero))
% 1.75/0.58  = { by axiom 6 (isolation) R->L }
% 1.75/0.58    strong_iteration(X)
% 1.75/0.58  
% 1.75/0.58  Goal 1 (goals): strong_iteration(x0) = addition(multiplication(strong_iteration(x0), x0), one).
% 1.75/0.58  Proof:
% 1.75/0.58    strong_iteration(x0)
% 1.75/0.58  = { by lemma 8 R->L }
% 1.75/0.58    addition(multiplication(strong_iteration(x0), zero), star(x0))
% 1.75/0.58  = { by axiom 1 (left_annihilation) R->L }
% 1.75/0.58    addition(multiplication(strong_iteration(x0), multiplication(zero, x0)), star(x0))
% 1.75/0.58  = { by axiom 2 (additive_commutativity) R->L }
% 1.75/0.58    addition(star(x0), multiplication(strong_iteration(x0), multiplication(zero, x0)))
% 1.75/0.58  = { by axiom 5 (star_unfold2) R->L }
% 1.75/0.58    addition(addition(one, multiplication(star(x0), x0)), multiplication(strong_iteration(x0), multiplication(zero, x0)))
% 1.75/0.58  = { by axiom 4 (additive_associativity) R->L }
% 1.75/0.58    addition(one, addition(multiplication(star(x0), x0), multiplication(strong_iteration(x0), multiplication(zero, x0))))
% 1.75/0.58  = { by axiom 3 (multiplicative_associativity) }
% 1.75/0.58    addition(one, addition(multiplication(star(x0), x0), multiplication(multiplication(strong_iteration(x0), zero), x0)))
% 1.75/0.58  = { by axiom 7 (distributivity2) R->L }
% 1.75/0.58    addition(one, multiplication(addition(star(x0), multiplication(strong_iteration(x0), zero)), x0))
% 1.75/0.58  = { by axiom 2 (additive_commutativity) R->L }
% 1.75/0.58    addition(one, multiplication(addition(multiplication(strong_iteration(x0), zero), star(x0)), x0))
% 1.75/0.58  = { by axiom 2 (additive_commutativity) R->L }
% 1.75/0.58    addition(multiplication(addition(multiplication(strong_iteration(x0), zero), star(x0)), x0), one)
% 1.75/0.58  = { by lemma 8 }
% 1.75/0.58    addition(multiplication(strong_iteration(x0), x0), one)
% 1.75/0.58  % SZS output end Proof
% 1.75/0.58  
% 1.75/0.58  RESULT: Theorem (the conjecture is true).
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