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Proof step 4 from proof put-at (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a]
Proof step 6:
m ≤ # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ (a0 ' + x0)[m, a][n] = (a0 ' + x0)[n], m = n, m ≠ 0 ∧ a0 ' + x0 ≠ []
Proof step 11:
m ≤ # x0, ∀ m, a, n. m < # x0 → x0[m, a][n] = x0[n] ∨ m = n ⊦ a0 ' + x0 = [], ((a0 ' + x0).head ' + (a0 ' + x0).tail[m -1, a])[n] = (a0 ' + x0)[n], m = n, m = 0

Rule argument

spec: <toplevel> name: put-rec seq: ⊦ n ≠ 0 ∧ x ≠ [] → x[n, a] = x.head ' + x.tail[n -1, a] subst: [a, n, x] → [a, m, a0 ' + x0] flags: <all paths>

Additional information

Used terms: a, m, a0 ' + x0