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Proof step 10 from proof firstN-lastN (previous step)

1 + # x0 = n0 + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ firstn(n0, a ' + x0) + restn(n0, a ' + x0) = a ' + x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 12:
1 + # x0 = n0 + n, ∀ n0, n. # x0 = n0 + n → firstn(n0, x0) + restn(n0, x0) = x0 ⊦ (a ' + firstn(n0 -1, x0)) + restn(n0, a ' + x0) = a ' + x0, n0 = 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] → [x0, a, n0] flags: <all paths>

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%restN-zero: ⊦ restn(0, x) = x;% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s) Used terms: x0, a, n0