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Proof step 8 from proof s2n-app (previous step)

∀ str0. str0 = "" ∨ s→n(str2 + str0) = s→n(str2) * 10 ^ # str0 + s→n(str0) ⊦ s→n(char0 ') * 10 ^ # str2 * 10 ^ # str0 + s→n(str2 + str0) = (10 ^ # str0 * s→n(char0 ') * 10 ^ # str2 + 10 ^ # str0 * s→n(str2)) + s→n(str0), str0 = "", char0 ' + str2 = "", str2 = ""

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ str0. str0 = "" ∨ s→n(str2 + str0) = s→n(str2) * 10 ^ # str0 + s→n(str0) ⊦ str0 = "", char0 ' + str2 = "", str2 = "", s→n(str2 + str0) = 10 ^ # str0 * s→n(str2) + s→n(str0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult