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Proof step 1 from proof l2s-app (conclusion)

⊦ l→s(chars + chars0) = l→s(chars) + l→s(chars0)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ chars0. l→s(chars1 + chars0) = l→s(chars1) + l→s(chars0) ⊦ l→s((char + chars1) + chars0) = l→s(char + chars1) + l→s(chars0)

Rule argument

indvar: chars

Additional information

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