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Proof step 3 from proof rest-in (previous step)

a ∈ restn(n, x), n ≤ # x ⊦ a ∈ x

User applied rule structural induction

And obtained the following new goals

Proof step 4:
a ∈ restn(n0 +1, x), n0 +1 ≤ # x, ∀ a, x. a ∈ restn(n0, x) ∧ n0 ≤ # x → a ∈ x ⊦ a ∈ x

Rule argument

indvar: n

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%nat freely generated by 0, +1;;% ⊦ 0 ≤ m; (s)