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

⊦ H0 ⊥ (H1 ++ r)[r, o] ↔ H0 ⊥ H1 ∧ ¬ r ∈ H0, r = null, r ∈ H1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)
Proof step 5:
⊦ (∀ r0. ¬ r0 ∈ H0 ∨ ¬ r0 ∈ (H1 ++ r)[r, o]) ↔ H0 ⊥ H1 ∧ ¬ r ∈ H0, r = null, r ∈ H1

Rule argument

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

Additional information

Used terms: (H1 ++ r)[r, o], H0