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Proof step 2 from proof s2n-rec-01 (previous step)

⊦ s→n(str) = s→n(str.rstring) + 10 ^ (# str - 1) * s→n(str.char1 '), # str = 1, str = ""

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
str = char + str0 ⊦ s→n(str) = s→n(str0) + 10 ^ (# str - 1) * s→n(char '), # str = 1, str = ""

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char0, str0] → [str, char0, str0] flags:

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str, char0, str0