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Proof step 9 from proof in-put (previous step)

a ∈ x0, m ≤ # x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ #oc(a, x0) = 1, a = a0, a ∈ (a0 ' + x0)[m, b]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b]
Proof step 11:
a ∈ x0, m ≤ # x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ #oc(a, x0) = 1, a = a0, m = 0, a ∈ a0 ' + x0[m -1, b]

Rule argument

spec: list-del name: put-rec-02 seq: ⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b] subst: [b, n, x, a] → [b, m, x0, a0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ (b ' + x)[0, a] = a + x; (s)% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss) Used terms: b, m, x0, a0