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Proof step 41 from proof precise-sep (previous step)

H4 ∪ H1 ∪ H0 = H4 ∪ H3 ∪ H, P0(H4), P1(H3), P1(H1), H4 ⊥ H3, H0 ⊥ H4, H4 ⊥ H, H ⊥ H3, H1 ⊥ H4, H0 ⊥ H1, ∀ H, H0, H1, H2. P1(H) ∧ P1(H0) ∧ H ⊥ H1 ∧ H0 ⊥ H2 ∧ H ∪ H1 = H0 ∪ H2 → H = H0 ⊦ H1 = H3, H4 ⊥ H3 ∪ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H)

Rule argument

spec: heap name: disj-distr-uni seq: ⊦ H1 ⊥ H → (H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H) subst: [H, H1, H0] → [H, H3, H4] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H, H3, H4