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Proof step 5 from proof notin-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ y, a ∈ filter(x0, y), ∀ a, y. a ∈ filter(x0, y) → a ∈ y ⊦ a ∈ y

Rule argument

Additional information

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