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Proof step 5 from proof in-01-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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