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Proof step 8 from proof restn-less (previous step)

n0 ≤ # y0, ∀ y, a. n0 ≤ # y → restn(n0, a ' + y) = (a ' + y)[n0] ' + restn(n0, y) ⊦ restn(n0, a0 ' + y0) = (a0 ' + y0)[n0] ' + restn(n0, y0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: y0, a0

Additional information

Used terms: y0, a0