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Proof step 1 from proof s2n-rec-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ s→n(str) = s→n(str.rstring) + 10 ^ (# str - 1) * s→n(str.char1 '), # str = 1, str = ""

Rule argument

Additional information

Used simplifier rules: ⊦ # str = 0 ↔ str = "";% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ str1 + "" = str1;% ⊦ char + str = char ' + str;% ⊦ n -1 = n - 1;