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Proof step 1 from proof sublist-append-02 (conclusion)

⊦ # x ≤ m → sublist(0, m, x + y) = x + sublist(0, m - # x, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# x ≤ m ⊦ firstn(m, x + y) = x + firstn(m - # x, y)

Rule argument

Additional information

Used simplifier rules: sublist-def: ⊦ sublist(m, n, x) = firstn(n, restn(m, x));%restN-zero: ⊦ restn(0, x) = x;