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The language of Presburger arithmetic extends predicate calculus with a constant 0, unary operator S, binary operator +, and relation =. Its axioms are: * \(\forall n: S(n) eq 0\) * \(\forall n \forall m: S(n) = S(m) \Rightarrow n = m\) * \(\forall n: n + 0 = n\) * \(\forall n \forall m: n + S(m) = S(n + m)\) * For every first-order formula \(\varphi(n)\): \((\varphi(0) \wedge (\forall n: \varphi(n) \Rightarrow \varphi(S(n)))) \Rightarrow \forall m: \varphi(m)\).

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  • Presburger arithmetic
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  • The language of Presburger arithmetic extends predicate calculus with a constant 0, unary operator S, binary operator +, and relation =. Its axioms are: * \(\forall n: S(n) eq 0\) * \(\forall n \forall m: S(n) = S(m) \Rightarrow n = m\) * \(\forall n: n + 0 = n\) * \(\forall n \forall m: n + S(m) = S(n + m)\) * For every first-order formula \(\varphi(n)\): \((\varphi(0) \wedge (\forall n: \varphi(n) \Rightarrow \varphi(S(n)))) \Rightarrow \forall m: \varphi(m)\).
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abstract
  • The language of Presburger arithmetic extends predicate calculus with a constant 0, unary operator S, binary operator +, and relation =. Its axioms are: * \(\forall n: S(n) eq 0\) * \(\forall n \forall m: S(n) = S(m) \Rightarrow n = m\) * \(\forall n: n + 0 = n\) * \(\forall n \forall m: n + S(m) = S(n + m)\) * For every first-order formula \(\varphi(n)\): \((\varphi(0) \wedge (\forall n: \varphi(n) \Rightarrow \varphi(S(n)))) \Rightarrow \forall m: \varphi(m)\).
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