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Proof step 3 from proof restN-append (previous step)

n ≤ # x ⊦ restn(n, x + y) = restn(n, x) + y

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n0 +1 ≤ # x, ∀ x, y. n0 ≤ # x → restn(n0, x + y) = restn(n0, x) + y ⊦ restn(n0 +1, x + y) = restn(n0 +1, x) + y

Rule argument

indvar: n

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%nat freely generated by 0, +1;;% ⊦ x + y = x + z ↔ y = z; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list