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Proof step 3 from proof subset-rec (previous step)

⊦ (∀ a0. a0 ∈ a ' + x → a0 ∈ y) ↔ a ∈ y ∧ x ⊆ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y) ↔ a ∈ y ∧ x ⊆ y

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