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Proof step 9 from proof prefix-num (previous step)

⊦ char ' + str1 = "", char ' + str2 = "", s→l(char ' + str1) ⊑ s→l(char ' + str2) ↔ str1 ⊑ str2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ char ' + str1 = "", char ' + str2 = "", s→l(str1) ⊑ s→l(str2) ↔ str1 ⊑ str2

Rule argument

Additional information

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