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Proof step 1 from proof sublist-restn-01 (conclusion)

⊦ m + n + m0 ≤ # x → sublist(m, n, restn(m0, x)) = sublist(m + m0, n, x)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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