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Proof step 5 from proof isprefix-append-size (previous step)

# x1 + 1 < # x0, ∃ z. x0 + z = x1 + a ' ⊦ x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x0 + z = x1 + a ', # x1 + 1 < # x0, a ∈ x0 ∨ a ∈ z ⊦ x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list