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Proof step 11 from proof substring-app-rec (previous step)

∀ str0, n0, n. substring(str1 + str0, n0 + # str1, n) = substring(str0, n0, n) ⊦ substring(str1 + str0, n0 + # str1, n) = substring(str0, n0, n), (char ' + str1) + str0 = ""

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: str0, n0, n

Additional information

Used terms: str0, n0, n