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Proof step 4 from proof put-del1-in (previous step)

m ≤ # x0, ∀ m, a. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a ∈ (a0 ' + x0)[m, a] -1l a ↔ a ∈ a0 ' + x0, (a0 ' + x0)[m] = a

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. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a0 ' + x0 = [], a ∈ ((a0 ' + x0).head ' + (a0 ' + x0).tail[m -1, a]) -1l a ↔ a ∈ a0 ' + x0, (a0 ' + x0)[m] = a, 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 simplifier rules: put-zero: ⊦ (b ' + x)[0, a] = a + x;%del1-y: ⊦ (a ' + x) -1l a = x;%get-zero-01: ⊦ (a ' + x)[0] = a;% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s) Used terms: a, m, a0 ' + x0