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Proof step 21 from proof map-putex (previous step)

P(H0), r ∈ ∅ ++ r ⊦ r ∈ H0, ∃ H2, H1. H2 ⊥ H1 ∧ (∅ ++ r)[r, o1] ∪ H0 = H2 ∪ H1 ∧ ((r => o1))(H2) ∧ P(H1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
P(H0), r ∈ ∅ ++ r ⊦ r ∈ H0, ∃ H1. ¬ r ∈ H1 ∧ (∅ ++ r)[r, o1] ∪ H0 = (∅ ++ r)[r, o1] ∪ H1 ∧ P(H1)

Rule argument

Additional information

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