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Proof step 15 from proof put-firstn-restn (previous step)

# y0 = m + 1, m ≤ # y0 ⊦ (y0 + z)[m, a] = firstn(m, y0) + a ' + z, y0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# y0 = m + 1 ⊦ (y0 + z)[m, a] = firstn(m, y0) + a ' + z, y0 = []

Rule argument

Additional information

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