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Proof step 30 from proof precise-ls (previous step)

(H4 ∪ H3) ∪ H0 = (H5 ∪ H2) ∪ H, ((r => a × r1))(H5), ((r => a × r0))(H4), ls(r1, x0)(H2), ls(r0, x0)(H3), H ⊥ H2, H ⊥ H5, H5 ⊥ H2, H0 ⊥ H3, H0 ⊥ H4, H4 ⊥ 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 = H5 ∪ H2, H4 = H5 ∧ r0 = r1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 32:
((r => a × r0))(H4), ((r => a × r1))(H5), ls(r0, x0)(H3), ls(r1, x0)(H2), H4 ⊥ H3, H0 ⊥ H4, H0 ⊥ H3, H5 ⊥ H2, H ⊥ H5, H ⊥ H2, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1, ∀ r. (r ∈ (H4 ∪ H3) ∪ H0 ↔ r ∈ (H5 ∪ H2) ∪ H) ∧ (r ∈ (H4 ∪ H3) ∪ H0 → ((H4 ∪ H3) ∪ H0)[r] = ((H5 ∪ H2) ∪ H)[r]) ⊦ H4 ∪ H3 = H5 ∪ H2, H4 = H5 ∧ r0 = r1

Rule argument

spec: heap, inst: cell-heap name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [(H5 ∪ H2) ∪ H, (H4 ∪ H3) ∪ H0] flags: <all paths>

Additional information

Used terms: (H5 ∪ H2) ∪ H, (H4 ∪ H3) ∪ H0