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Proof step 1 from proof firstn-firstn (conclusion)

⊦ m ≤ n ∧ n ≤ # x → firstn(m, firstn(n, x)) = firstn(m, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat