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Proof step 31 from proof precise-disj (previous step)

H ∪ H5 ∪ H0 = (H3 ∪ H2) ∪ H1, P0(H), P0(H3), P1(H2), P1(H5), precise(P1), H3 ⊥ H1, H2 ⊥ H3, H2 ⊥ H1, H5 ⊥ H0, H ⊥ H5, H ⊥ H0, ∀ H, H0, H1, H2. P0(H) ∧ P0(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ (H3 ∪ H2) ∪ H1 = H3 ∪ H2 ∪ H1, H ⊥ H2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H

Rule argument

spec: heap name: uni-ass seq: ⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H subst: [H, H1, H0] → [H1, H2, H3] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H1, H2, H3