TSTP Solution File: KLE083+1 by Twee---2.4.2

View Problem - Process Solution

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% File     : Twee---2.4.2
% Problem  : KLE083+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 : n010.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 05:35:49 EDT 2023

% Result   : Theorem 0.19s 0.41s
% Output   : Proof 0.19s
% Verified : 
% SZS Type : -

% Comments : 
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%----WARNING: Could not form TPTP format derivation
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%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12  % Problem  : KLE083+1 : TPTP v8.1.2. Released v4.0.0.
% 0.00/0.13  % Command  : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.12/0.34  % Computer : n010.cluster.edu
% 0.12/0.34  % Model    : x86_64 x86_64
% 0.12/0.34  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.34  % Memory   : 8042.1875MB
% 0.12/0.34  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.34  % CPULimit : 300
% 0.12/0.34  % WCLimit  : 300
% 0.12/0.34  % DateTime : Tue Aug 29 12:15:48 EDT 2023
% 0.12/0.34  % CPUTime  : 
% 0.19/0.41  Command-line arguments: --lhs-weight 1 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10
% 0.19/0.41  
% 0.19/0.41  % SZS status Theorem
% 0.19/0.41  
% 0.19/0.41  % SZS output start Proof
% 0.19/0.41  Take the following subset of the input axioms:
% 0.19/0.41    fof(additive_identity, axiom, ![A]: addition(A, zero)=A).
% 0.19/0.41    fof(domain1, axiom, ![X0]: multiplication(antidomain(X0), X0)=zero).
% 0.19/0.41    fof(domain3, axiom, ![X0_2]: addition(antidomain(antidomain(X0_2)), antidomain(X0_2))=one).
% 0.19/0.41    fof(domain4, axiom, ![X0_2]: domain(X0_2)=antidomain(antidomain(X0_2))).
% 0.19/0.41    fof(goals, conjecture, ![X0_2]: X0_2=multiplication(domain(X0_2), X0_2)).
% 0.19/0.41    fof(left_distributivity, axiom, ![B, C, A2]: multiplication(addition(A2, B), C)=addition(multiplication(A2, C), multiplication(B, C))).
% 0.19/0.41    fof(multiplicative_left_identity, axiom, ![A2]: multiplication(one, A2)=A2).
% 0.19/0.41  
% 0.19/0.41  Now clausify the problem and encode Horn clauses using encoding 3 of
% 0.19/0.41  http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 0.19/0.41  We repeatedly replace C & s=t => u=v by the two clauses:
% 0.19/0.41    fresh(y, y, x1...xn) = u
% 0.19/0.41    C => fresh(s, t, x1...xn) = v
% 0.19/0.41  where fresh is a fresh function symbol and x1..xn are the free
% 0.19/0.41  variables of u and v.
% 0.19/0.41  A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 0.19/0.41  input problem has no model of domain size 1).
% 0.19/0.41  
% 0.19/0.41  The encoding turns the above axioms into the following unit equations and goals:
% 0.19/0.41  
% 0.19/0.41  Axiom 1 (multiplicative_left_identity): multiplication(one, X) = X.
% 0.19/0.41  Axiom 2 (additive_identity): addition(X, zero) = X.
% 0.19/0.41  Axiom 3 (domain4): domain(X) = antidomain(antidomain(X)).
% 0.19/0.41  Axiom 4 (domain1): multiplication(antidomain(X), X) = zero.
% 0.19/0.41  Axiom 5 (domain3): addition(antidomain(antidomain(X)), antidomain(X)) = one.
% 0.19/0.41  Axiom 6 (left_distributivity): multiplication(addition(X, Y), Z) = addition(multiplication(X, Z), multiplication(Y, Z)).
% 0.19/0.41  
% 0.19/0.41  Goal 1 (goals): x0 = multiplication(domain(x0), x0).
% 0.19/0.41  Proof:
% 0.19/0.41    x0
% 0.19/0.41  = { by axiom 1 (multiplicative_left_identity) R->L }
% 0.19/0.41    multiplication(one, x0)
% 0.19/0.41  = { by axiom 5 (domain3) R->L }
% 0.19/0.41    multiplication(addition(antidomain(antidomain(x0)), antidomain(x0)), x0)
% 0.19/0.41  = { by axiom 3 (domain4) R->L }
% 0.19/0.41    multiplication(addition(domain(x0), antidomain(x0)), x0)
% 0.19/0.41  = { by axiom 6 (left_distributivity) }
% 0.19/0.41    addition(multiplication(domain(x0), x0), multiplication(antidomain(x0), x0))
% 0.19/0.41  = { by axiom 4 (domain1) }
% 0.19/0.41    addition(multiplication(domain(x0), x0), zero)
% 0.19/0.41  = { by axiom 2 (additive_identity) }
% 0.19/0.41    multiplication(domain(x0), x0)
% 0.19/0.41  % SZS output end Proof
% 0.19/0.41  
% 0.19/0.41  RESULT: Theorem (the conjecture is true).
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