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Proof step 4 from proof in-01-01 (previous step)

a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 + x0, y) ↔ a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a ∈ a0 ' + x0

Rule argument

Additional information

Used simplifier rules: ⊦ a + x = a ' + x; (s)