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Proof step 3 from proof map-write (previous step)

((r => o0 * P))(H[r, o1]), r ∈ H ⊦ o0 = o1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 5:
r ∈ H, ∃ H0, H1. H0 ⊥ H1 ∧ H[r, o1] = H0 ∪ H1 ∧ ((r => o0))(H0) ∧ P(H1) ⊦ o0 = o1

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] → [H[r, o1], P, r => o0] flags: <all paths>

Additional information

Used terms: H[r, o1], P, r => o0