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Proof step 2 from proof l2s-app (previous step)

∀ chars0. l→s(chars1 + chars0) = l→s(chars1) + l→s(chars0) ⊦ l→s((char + chars1) + chars0) = l→s(char + chars1) + l→s(chars0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l2s-rec: ⊦ l→s(char ' + chars) = char + l→s(chars);% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ str + str0 = str + str1 ↔ str0 = str1;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char + str = char ' + str;% ⊦ char + chars = char ' + chars;