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Proof step 7 from proof del-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x)
Proof step 9:
¬ a0 ≠ a, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), perm(y, x), a0 ∈ x -1l a ∨ a0 ∈ y -1l a ⊦ #oc(a0, x -1l a) = #oc(a0, y -1l a)
Proof step 23:
a0 ≠ a, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y), perm(y, x), a0 ∈ x -1l a ∨ a0 ∈ y -1l a ⊦ #oc(a0, x) = #oc(a0, y -1l a)

Rule argument

spec: inst: name: del1-01 seq: ⊦ a ≠ b → #oc(a, x -1l b) = #oc(a, x) subst: [b, x, a] → [a, x, a0] flags: rewrite, current

Additional information

Used terms: a, x, a0