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Proof step 46 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, P(mk(rs2, f)), r ∈ rs2, r ∈ rs2 ∪ rs1, ∀ r. r ∈ rs2 → g(r) = a, ∀ r. (g(r) = a ∧ r ∈ rs2 ∪ rs1 ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs2 ∪ rs1 → f(r) = f0(r)) ⊦ # rs2 = 0, # rs2 + # rs1 = 0, g(r) = a, g0(r) = a, rs2 = ∅, rs2 ∪ rs1 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r

Additional information

Used terms: r