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Proof step 3 from proof disj-alloc (previous step)

H0 ⊥ H1 ⊦ r0 = null, r0 ∈ H1, r0 ∈ H0, H0 ⊥ H1 ++ r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)

Rule argument

spec: <toplevel> name: disj seq: ⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) subst: [H1, H0] → [H1 ++ r0, H0] flags: <all paths>

Additional information

Used simplifier rules: in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H;%disj-l: ⊦ r ∈ H0 ∧ H0 ⊥ H1 → (r ∈ H1 ↔ false); Used terms: H1 ++ r0, H0