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Proof step 8 from proof map-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
((r => o))(H0), P(H), H0 ⊥ H, ∀ r0. (r0 ∈ rs ↔ r0 ∈ H0 ∨ r0 ∈ H) ∧ (r0 ∈ rs → f(r0) = (H0 ∪ H)[r0]) ⊦ r ∈ rs

Rule argument

Additional information

Used simplifier rules: in-mk: ⊦ r ∈ mk(rs, f) ↔ r ∈ rs;%at-mk: ⊦ r ∈ rs → mk(rs, f)[r] = f(r);% ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)