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Proof step 9 from proof map-in (previous step)

((r => o))(H0), P(H), H0 ⊥ H, ∀ r0. (r0 ∈ rs ↔ r0 ∈ H0 ∨ r0 ∈ H) ∧ (r0 ∈ rs → f(r0) = (H0 ∪ H)[r0]) ⊦ r ∈ rs

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: r

Additional information

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