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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a]
Proof step 5:
¬(m ≠ 0 ∧ a0 ' + x0 ≠ []), m ≤ # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m = n, (a0 ' + x0)[m, a][n] = (a0 ' + x0)[n]
Proof step 10:
m ≠ 0 ∧ a0 ' + x0 ≠ [], m ≤ # x0, ∀ m, n, a. m < # x0 → m ≠ n → x0[m, a][n] = x0[n] ⊦ m = n, ((a0 ' + x0).first ' + (a0 ' + x0).rest[m -1, a])[n] = (a0 ' + x0)[n]

Rule argument

spec: list-del inst: name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.first ' + x.rest[n -1, a] subst: [a, n, x] → [a, m, a0 ' + x0] flags: rewrite, current

Additional information

Used terms: a, m, a0 ' + x0