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Proof step 3 from proof put-at-eq (previous step)

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

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:
m ≠ 0, m ≤ # x0, ∀ m, a. m < # x0 → x0[m, a][m] = a ⊦ (a0 ' + x0[m -1, a])[m] = a

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, m, x0, a0] flags: rewrite, current

Additional information

Used simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%get-zero-01: ⊦ (a ' + x)[0] = a;%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a, m, x0, a0