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Proof step 6 from proof disj-write (previous step)

∀ r0. ¬ r0 ∈ H0 ∨ r0 ≠ r ∧ ¬ r0 ∈ H1 ⊦ r = null, r ∈ H1, H0 ⊥ H1 ∧ ¬ r ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)
Proof step 8:
∀ r0. ¬ r0 ∈ H0 ∨ r0 ≠ r ∧ ¬ r0 ∈ H1 ⊦ r = null, r ∈ H1, (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) ∧ ¬ r ∈ H0

Rule argument

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

Additional information

Used terms: H1, H0