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Proof step 13 from proof exin-uni (previous step)

⊦ (H0 ++ r)[r, o] = H0 ∪ (∅ ++ r)[r, o], r = null, r ∈ H0, H0 ⊥ ∅ ++ r

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)

Rule argument

spec: <toplevel> name: disj seq: ⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) subst: [H1, H0] → [∅ ++ r, H0] flags: <all paths>

Additional information

Used simplifier rules: in-emp: ⊦ ¬ r ∈ ∅;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; Used terms: ∅ ++ r, H0