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Proof step 8 from proof ex-shift-o (previous step)

P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ∃ o. ((!(me, O, oP(o)) * P))(H1 ∪ H0)

User applied rule exists right

And obtained the following new goals

Proof step 9:
P(H0), oP(o)(H1), H1 ⊥ H0, ∀ r. r ∈ H1 → O(r) = me ⊦ ((!(me, O, oP(o)) * P))(H1 ∪ H0), ∃ o. ((!(me, O, oP(o)) * P))(H1 ∪ H0)

Rule argument

pos: right 1 terms: o abort: F discard: F computed: T

Additional information

Used terms: o