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Proof step 10 from proof lift-mapex (previous step)

((r => o))(H) ⊦ (∃ o. ((r => o))(H)) ∧ (r0 ∈ H → (r = r0 ⊃ me;O(r0)) = me)

User applied rule case distinction

And obtained the following new goals

Proof step 11:
((r => o))(H) ⊦ ∃ o. ((r => o))(H)
Proof step 12:
((r => o))(H), r0 ∈ H ⊦ (r = r0 ⊃ me;O(r0)) = me

Rule argument

pos: right 1

Additional information