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Proof step 9 from proof precise-ls (previous step)

(H4 ∪ H3) ∪ H0 = H1 ∪ H, ((r => a × r1 * ls(r1, x0)))(H1), ((r => a × r0))(H4), ls(r0, x0)(H3), H1 ⊥ H, H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1 ⊦ H4 ∪ H3 = H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 11:
(H4 ∪ H3) ∪ H0 = H1 ∪ H, ((r => a × r0))(H4), ls(r0, x0)(H3), H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ H3, H1 ⊥ H, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1, ∃ H0, H2. H0 ⊥ H2 ∧ H1 = H0 ∪ H2 ∧ ((r => a × r1))(H0) ∧ ls(r1, x0)(H2) ⊦ H4 ∪ H3 = 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] → [H1, ls(r1, x0), r => a × r1] flags: <all paths>

Additional information

Used terms: H1, ls(r1, x0), r => a × r1