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Proof step 15 from proof map-putex (previous step)

∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ ((r => o))(H0) ∧ P(H1) ⊦ ((r => o1 * P))(H[r, o1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
P(H0) ⊦ r = null, ((r => o1 * P))(((∅ ++ r)[r, o] ∪ H0)[r, o1]), r ∈ H0

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)