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Proof step 6 from proof in-01-01 (previous step)

a ∈ y, a0 ∈ 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 7:
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);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list