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Proof step 2 from proof append-eqlen (previous step)

# str = # str1 ⊦ str + str0 = str1 + str2 ↔ str = str1 ∧ str0 = str2

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string]%⊦ # str = # str0 → (str + str1 = str0 + str2 ↔ str = str0 ∧ str1 = str2); (ws)