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Proof step 3 from proof del1 (previous step)

a ∈ x ⊦ #oc(a, x -1l a) = #oc(a, x) - 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
#oc(a, x) = m + 1, a ∈ x ⊦ #oc(a, x -1l a) = m, #oc(a, x) = 0

Rule argument

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

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, x), m