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Proof step 3 from proof restn-first-get (previous step)

n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ restn(n, a ' + x0).first = (a ' + x0)[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < n → restn(n, x) = restn(n -1, x.rest)
Proof step 5:
0 < n, n ≤ # x0, ∀ n. n < # x0 → restn(n, x0).first = x0[n] ⊦ restn(n -1, (a ' + x0).rest).first = (a ' + x0)[n]

Rule argument

spec: list-del inst: name: rest-rec seq: ⊦ 0 < n → restn(n, x) = restn(n -1, x.rest) subst: [x, n] → [a ' + x0, n] flags: rewrite, current

Additional information

Used simplifier rules: restN-zero: ⊦ restn(0, x) = x;%get-zero-01: ⊦ (a ' + x)[0] = a;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%c-08: ⊦ (a ' + x).first = a; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a ' + x0, n