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Proof step 11 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, m = 0, a ∈ x0[m - 1, b]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
m = m1 + 1, 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, m = 0, a ∈ x0[m1, b]

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m1] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m1