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Proof step 2 from proof string2string (previous step)

l→s(s→l(stringvar)) = stringvar ⊦ l→s(s→l(char0 + stringvar)) = char0 + stringvar

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: s2l-rec: ⊦ s→l(char ' + str) = char ' + s→l(str);%l2s-app: ⊦ l→s(chars + chars0) = l→s(chars) + l→s(chars0);%l2s-one: ⊦ l→s(char ') = char ';% ⊦ str + str0 = str + str1 ↔ str0 = str1;% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;% ⊦ char + str = char ' + str;