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Proof step 8 from proof in-put (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ n ≠ 0 → (a ' + x)[n, b] = a ' + x[n -1, b]
Proof step 10:
m ≠ 0, m ≤ # x0, a ∈ x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ a = a0, #oc(a, x0) = 1, a ∈ a0 ' + x0[m -1, b]

Rule argument

spec: list-del inst: 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: rewrite

Additional information

Used simplifier rules: lez: ⊦ 0 ≤ m; from specification nat%put-zero: ⊦ (b ' + x)[0, a] = a + x; from specification list-del%c: ⊦ a + x = a ' + x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-12: ⊦ a ∈ y → a ∈ x + y; from specification list Used terms: b, m, x0, a0