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Proof step 21 from proof del-04 (previous step)

#oc(a, x) = m + 1, perm(y, x), a ∈ y, a ∈ x, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), a ∈ x -1l a ∨ a ∈ y -1l a ⊦ m = #oc(a, y) - 1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 22:
#oc(a, x) = m + 1, #oc(a, y) = m0 + 1, perm(y, x), a ∈ x, a ∈ y, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), a ∈ x -1l a ∨ a ∈ y -1l a ⊦ #oc(a, y) = 0, m = m0

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [#oc(a, y), m0]

Additional information

Used simplifier rules: in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #oc(a, y), m0