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Proof step 19 from proof map-putex (previous step)

P(H0), r ∈ ∅ ++ r ⊦ ((r => o1 * P))((∅ ++ r)[r, o1] ∪ H0), r ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 21:
P(H0), r ∈ ∅ ++ r ⊦ r ∈ H0, ∃ H2, H1. H2 ⊥ H1 ∧ (∅ ++ r)[r, o1] ∪ H0 = H2 ∪ H1 ∧ ((r => o1))(H2) ∧ P(H1)

Rule argument

spec: heap-sep name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [(∅ ++ r)[r, o1] ∪ H0, P, r => o1] flags: <all paths>

Additional information

Used terms: (∅ ++ r)[r, o1] ∪ H0, P, r => o1