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Proof step 3 from proof in-01-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ y, ∀ a, y. a ∈ y → (a ∈ filter(x0, y) ↔ a ∈ x0) ⊦ a ∈ filter(a0 ' + x0, y) ↔ a = a0 ∨ a ∈ x0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list