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Proof step 2 from proof s2l-app (previous step)

∀ str0. s→l(str1 + str0) = s→l(str1) + s→l(str0) ⊦ s→l((char + str1) + str0) = s→l(char + str1) + s→l(str0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ str0. s→l(str1 + str0) = s→l(str1) + s→l(str0) ⊦ s→l(str1 + str0) = s→l(str1) + s→l(str0)

Rule argument

Additional information

Used simplifier rules: s2l-rec: ⊦ s→l(char ' + str) = char ' + s→l(str);% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ chars + chars0 = chars + chars1 ↔ chars0 = chars1;% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ char + str = char ' + str;