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Proof step 8 from proof Sep-eq (previous step)

((aP ⊗_ s))(H) → ((aP0 ⊗_ s))(H), ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ ((aP ⊗_ s))(H) ↔ ((aP0 ⊗_ s))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
((aP0 ⊗_ s))(H), ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ ((aP ⊗_ s))(H)

Rule argument

Additional information