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Proof step 2 from proof first (previous step)

str ≠ "" ⊦ (str + str0).char1 = str.char1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: first-01: ⊦ str ≠ "" → (str + str0).char1 = str.char1;