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

firstn(n, a ' + y) = a0 ' + y0, m0 ≤ # y, n ≤ # y + 1, m0 < n, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(m0, y0) = firstn(m0, y), n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 10:
n ≠ 0, a ' + firstn(n -1, y) = a0 ' + y0, m0 ≤ # y, n ≤ # y + 1, m0 < n, ∀ n, x. m0 ≤ n ∧ n ≤ # x ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ firstn(m0, y0) = firstn(m0, y), n = 0

Rule argument

spec: list-del inst: name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [y, a, n] flags: rewrite, current

Additional information

Used terms: y, a, n