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Proof step 7 from proof del-05 (previous step)

#oc(a, x) = #oc(a, x) - 1, a ∈ x, perm(x -1l a, x), ∀ a0. a0 ∈ x ∨ a0 ∈ x -1l a → #oc(a0, x) = #oc(a0, x -1l a) ⊦

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: #oc(a, x), m