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Proof step 67 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, P(mk(rs2, f0)), r ∈ rs2, r ∈ rs2 ∪ rs1, ∀ r. r ∈ rs2 → g0(r) = a ⊦ # rs2 = 0, # rs2 + # rs1 = 0, g0(r) = a, rs2 = ∅, rs2 ∪ (rs \ rs2) = rs, rs2 ∪ rs1 = ∅, r ∈ rs

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r

Additional information

Used terms: r