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Proof step 1 from proof Imp-from-eq (conclusion)

⊦ P0 = P1 ↔ P0 ⊆ P1 ∧ P1 ⊆ P0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ H. P0(H) → P1(H), ∀ H. P1(H) → P0(H) ⊦ P0 = P1

Rule argument

Additional information

Used simplifier rules: Imp: ⊦ P0 ⊆ P1 ↔ (∀ H. P0(H) → P1(H));