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Proof step 4 from proof firstN-lastN-in (previous step)

n ≤ # x → firstn(# x - n, x) + lastn(n, x) = x, n ≤ # x, a ∈ x ⊦ a ∈ lastn(n, x), a ∈ firstn(# x - n, x)

User applied rule case distinction

And obtained the following new goals

Proof step 5:
firstn(# x - n, x) + lastn(n, x) = x, n ≤ # x, a ∈ x ⊦ a ∈ lastn(n, x), a ∈ firstn(# x - n, x)

Rule argument

pos: left 1

Additional information