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Proof step 3 from proof mapex-del (previous step)

((r => o * P))(H) ⊦ P(H -- r)

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))

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, P, r => o] flags: <all paths>

Additional information

Used simplifier rules: maplet: ⊦ ((r => o))(H) ↔ H = (∅ ++ r)[r, o] ∧ r ≠ null;% ⊦ H0 ⊥ H1 → (H0 ∪ H1) -- r = (H0 -- r) ∪ (H1 -- r); (s)%r ≠ null ⊦ (∅ ++ r)[r, o] ⊥ H ↔ ¬ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)%r ∈ H ⊦ H[r, o] -- r = H -- r; (s)%¬ r ∈ H, r ≠ null ⊦ (H ++ r) -- r = H; (s)% ⊦ ¬ r ∈ H → H -- r = H; (s)% ⊦ ∅ ∪ H = H; (s) Used terms: H, P, r => o