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Proof step 8 from proof restn-firstn (previous step)

m0 + n0 + m0 < # y, ∀ n0, x. m0 + n0 + m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, y)) = firstn(n0, restn(m0, y)), y = []

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: n0, y

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat Used terms: n0, y