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Proof step 24 from proof owner-lst1 (previous step)

rs1 ∩ rs0 = ∅, ((r => a × r0))(H0), R[rs0, cf, af], ls(r0, x0)(H), H0 ⊥ H, ∀ r1. (r1 ∈ rs1 ↔ r1 ∈ H0 ∨ r1 ∈ H) ∧ (r1 ∈ rs1 → cf(r1) = (H0 ∪ H)[r1]) ⊦ af(r) = o, r = null, r ∈ rs1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: r

Additional information

Used simplifier rules: ⊦ ((r => ce))(H) → r ∈ H; (s) Used terms: r