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Proof step 1 from proof rest-sub (conclusion)

⊦ n ≤ # x → restn(n, x) = sublist(n, # x - n, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n ≤ # x ⊦ restn(n, x) = firstn(# x - n, restn(n, x))

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));