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Proof step 3 from proof restn-firstn-better (previous step)

n = n0 + m, m ≤ n, n ≤ # x, m ≤ # x ⊦ restn(m, firstn(n, x)) = firstn(n0, restn(m, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 + m ≤ # x, m ≤ # x ⊦ restn(m, firstn(n0 + m, x)) = firstn(n0, restn(m, x))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat