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Proof step 11 from proof ex-shift-o-p (previous step)

P(H0), P0(H), oP(o)(H2), H2 ⊥ H, H0 ⊥ H2, H0 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0)

User applied rule exists right

And obtained the following new goals

Proof step 12:
P(H0), P0(H), oP(o)(H2), H0 ⊥ H, H0 ⊥ H2, H2 ⊥ H, ∀ r. r ∈ H2 ∨ r ∈ H → O(r) = me ⊦ ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0), ∃ o. ((!(me, O, oP(o) * P0) * P))((H2 ∪ H) ∪ H0)

Rule argument

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

Additional information

Used terms: o