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Proof step 14 from proof precise-ls (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H
Proof step 16:
H5 ∪ H3 ∪ H0 = (H5 ∪ H2) ∪ H, ((r => a × r1))(H5), ls(r1, x0)(H2), ls(r1, x0)(H3), H ⊥ H2, H ⊥ H5, H5 ⊥ H2, H0 ⊥ H3, H5 ⊥ H3, H3 ⊥ H0, H0 ⊥ H5, ∀ H2, H0, H1, H, r. H2 ∪ H0 = H1 ∪ H ∧ ls(r, x0)(H2) ∧ ls(r, x0)(H1) ∧ H2 ⊥ H0 ∧ H1 ⊥ H → H2 = H1 ⊦ H3 = H2

Rule argument

spec: heap, inst: cell-heap name: uni-ass seq: ⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H subst: [H, H1, H0] → [H0, H3, H5] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H0, H3, H5