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Proof step 9 from proof firstn-firstn (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 11:
a ' + firstn(n -1, y) = a0 ' + y0, m0 < n, n ≤ 1 + # y, m0 ≤ # y, ∀ n, x. n ≤ # x ∧ m0 ≤ n ∧ m0 ≤ # x → firstn(m0, firstn(n, x)) = firstn(m0, x) ⊦ a0 ' + firstn(m0, y0) = a ' + firstn(m0, y), n = 0

Rule argument

spec: <toplevel> name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [y, a, n] flags: <all paths>

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];% ⊦ x + y = [] ↔ x = [] ∧ y = []; (as)% ⊦ a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: y, a, n