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Proof step 8 from proof dropn-first (previous step)

⊦ y = [], z = (y + z = [] ⊃ [];restn(# y - 1, y.tail + z))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
⊦ y + z = [], y = [], z = restn(# y - 1, y.tail + z)

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)%a: ⊦ (x + y) + z = x + y + z; from specification list