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Proof step 64 from proof lift-unknown (previous step)

mk(rs1, f) = (∅ ++ r)[r, o], rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ r = null, ∃ o, a. r ==>(o, a)(mk(rs1, f, g)), ∃ o, a. (r ==>(o, a) * R)[rs1 ∪ rs0, f, g]

User applied rule exists right

And obtained the following new goals

Proof step 65:
mk(rs1, f) = (∅ ++ r)[r, o], rs1 ∩ rs0 = ∅, R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ r = null, r ==>(o, a)(mk(rs1, f, g)), ∃ o, a. (r ==>(o, a) * R)[rs1 ∪ rs0, f, g]

Rule argument

pos: right 2 terms: o, a abort: F discard: T computed: T

Additional information

Used terms: o, a