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Proof step 8 from proof oc-diff-01 (previous step)

∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, (a0 ' + x0) \ y) = #oc(a, x0), a ∈ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ y → (a ' + x) \ y = x \ y
Proof step 10:
¬ a0 ∈ y, ∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, (a0 ' + x0) \ y) = #oc(a, x0), a ∈ y
Proof step 12:
a0 ∈ y, ∀ a, y. #oc(a, x0 \ y) = #oc(a, x0) ∨ a ∈ y ⊦ a0 = a, #oc(a, x0 \ y) = #oc(a, x0), a ∈ y

Rule argument

spec: list-set inst: name: diff-y seq: ⊦ a ∈ y → (a ' + x) \ y = x \ y subst: [y, x, a] → [y, x0, a0] flags: rewrite

Additional information

Used terms: y, x0, a0