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

m ≤ # x0, ∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a = a0 ∨ a ∈ x0, a = (a0 ' + x0)[m]

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, ∀ a, m, b. a ≠ x0[m] ∧ m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0) ⊦ a ∈ (a0 ' + x0).first ' + (a0 ' + x0).rest[m -1, b] ↔ a = b ∨ a = a0 ∨ a ∈ x0, a = (a0 ' + x0)[m]

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] → [b, m, a0 ' + x0] flags: rewrite, current

Additional information

Used simplifier rules: get-zero-01: ⊦ (a ' + x)[0] = a;%put-zero: ⊦ (b ' + x)[0, a] = a + x;%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list Used terms: b, m, a0 ' + x0