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Proof step 1 from proof map-del (conclusion)

⊦ ((r => o * P))(H) ↔ P(H -- r) ∧ r ∈ H ∧ H[r] = o

User applied rule case distinction

And obtained the following new goals

Proof step 2:
((r => o * P))(H) ⊦ P(H -- r) ∧ r ∈ H ∧ H[r] = o
Proof step 5:
H[r] = o, P(H -- r), r ∈ H ⊦ ((r => o * P))(H)

Rule argument

pos: right 1

Additional information