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Proof step 3 from proof put-app-left (previous step)

n ≤ # x0, ∀ n, y, a. n < # x0 → (x0 + y)[n, a] = x0[n, a] + y ⊦ ((a0 ' + x0) + y)[n, a] = (a0 ' + x0)[n, a] + y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b]
Proof step 5:
n ≠ 0, n ≤ # x0, ∀ n, y, a. n < # x0 → (x0 + y)[n, a] = x0[n, a] + y ⊦ a0 ' + (x0 + y)[n -1, a] = (a0 ' + x0)[n, a] + y

Rule argument

spec: list-del inst: name: put-rec-02 seq: ⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b] subst: [b, n, x, a] → [a, n, x0 + y, a0] flags: rewrite, current

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%c: ⊦ a + x = a ' + x; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a, n, x0 + y, a0