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

P(H -- r), r ∈ H ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ ((r => H[r]))(H0) ∧ P(H1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
P(H -- r), r ∈ H ⊦ ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ H0 = (∅ ++ r)[r, (H0 ∪ H1)[r]] ∧ P(H1)

Rule argument

Additional information

Used simplifier rules: maplet: ⊦ ((r => o))(H) ↔ H = (∅ ++ r)[r, o] ∧ r ≠ null;% ⊦ r ∈ H → r ≠ null; (s)