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

a ∈ y ⊦ a ∈ filter(x, y) ↔ a ∈ x

User applied rule structural induction

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 + x0

Rule argument

indvar: x

Additional information

Used simplifier rules: filt-e: ⊦ filter([], y) = [];%list freely generated by [], +;;% ⊦ ¬ a ∈ []; (s)