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Proof step 1 from proof l2s-one (conclusion)

⊦ l→s(char ') = char '

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ l→s(char ' + chars) = char + l→s(chars)

Rule argument

spec: string2list inst: name: l2s-rec seq: ⊦ l→s(char ' + chars) = char + l→s(chars) subst: [char, chars] → [char, []] flags: current, []

Additional information

Used simplifier rules: l2s-empty: ⊦ l→s([]) = "";% ⊦ chars + [] = chars;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char + str = char ' + str;% ⊦ str1 + "" = str1;% ⊦ (str1 + str2) + str3 = str1 + str2 + str3; Used terms: char, []