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Proof step 3 from proof mapex-new (previous step)

⊦ P(H) ↔ (∃ o. ((r => o * P))(H ++ r)), r = null, r ∈ H

User applied rule case distinction

And obtained the following new goals

Proof step 4:
P(H) ⊦ r = null, r ∈ H, ∃ o. ((r => o * P))(H ++ r)
Proof step 21:
∃ o. ((r => o * P))(H ++ r) ⊦ r = null, P(H), r ∈ H

Rule argument

pos: right 1

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