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Proof step 9 from proof oc-filt (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ∈ y → filter(a ' + x, y) = a ' + filter(x, y)
Proof step 11:
a ∈ y, ∀ a, y. a ∈ y → #oc(a, filter(x0, y)) = #oc(a, x0) ⊦ #oc(a, filter(a0 ' + x0, y)) = #oc(a, x0), a0 = a, a0 ∈ y
Proof step 13:
a ∈ y, a0 ∈ y, ∀ a, y. a ∈ y → #oc(a, filter(x0, y)) = #oc(a, x0) ⊦ #oc(a, a0 ' + filter(x0, y)) = #oc(a, x0), a0 = a

Rule argument

spec: list-set name: filt-y seq: ⊦ a ∈ y → filter(a ' + x, y) = a ' + filter(x, y) subst: [y, x, a] → [y, x0, a0] flags: <all paths>

Additional information

Used terms: y, x0, a0