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Proof step 15 from proof firstn-append-01 (previous step)

# x0 = n1 + # z, n1 = m + 1, x0 = firstn(m, x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z ⊦ n1 = 0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z, x0 = firstn(n1, x0) + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# x0 = (m + 1) + # z, x0 = firstn(m, x0) + a ' + z, # z ≤ # x0, ∀ z, a. x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → x0 = firstn(# x0 - # z, x0) + z ⊦ x0 = firstn(# x0 - 1 + # z, x0) + a ' + z, x0 = (x0.head ' + firstn(m, x0.tail)) + z

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.head + firstn(n, x.tail);% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a + x = a ' + x; (s)