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Proof step 4 from proof restn-less (previous step)

n ≤ # y ⊦ restn(n, a ' + y) = (a ' + y)[n] ' + restn(n, y)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ y, a. n0 ≤ # y → restn(n0, a ' + y) = (a ' + y)[n0] ' + restn(n0, y), n0 +1 ≤ # y ⊦ restn(n0 +1, a ' + y) = (a ' + y)[n0 +1] ' + restn(n0 +1, y)

Rule argument

term: n

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%get-zero-01: ⊦ (a ' + x)[0] = a;%nat freely generated by 0, +1;;%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list