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Proof step 9 from proof sublist-last (previous step)

m0 + n0 < # y, ∀ m0, x. m0 + n0 < # x → (restn(m0, x).first ' + firstn(n0, restn(m0, x).rest)).last = x[m0 + n0] ∨ x = [] ⊦ y = [], (restn(m0, y).first ' + firstn(n0, restn(m0, y).rest)).last = y[m0 + n0]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: m0, y

Additional information

Used terms: m0, y