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Proof step 18 from proof del-04 (previous step)

#oc(a, x) = m + 1, perm(y, x), 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 -1l a)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1
Proof step 20:
#oc(a, x) = m + 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 ⊦ m = #oc(a, y) -1

Rule argument

spec: <toplevel> name: del1 seq: ⊦ a ∈ x → #oc(a, x -1l a) = #oc(a, x) -1 subst: [x, a] → [y, a] flags: <all paths>

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%in: ⊦ a ∈ y ∧ ¬ a ∈ x → ¬ perm(x, y);% ⊦ ¬ a ∈ x → x -1l a = x; (s) Used terms: y, a