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Proof step 19 from proof del-uni (previous step)

r0 ∈ H1, r0 ∈ (H0 -- r) ∪ (H1 -- r), H0 ⊥ H1 ⊦ H1[r0] = ((H0 -- r) ∪ (H1 -- r))[r0], r0 = r, H0 -- r ⊥ H1 -- r

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ 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 -- r, H0 -- r] flags: <all paths>

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;%disj-l: ⊦ r ∈ H0 ∧ H0 ⊥ H1 → (r ∈ H1 ↔ false); Used terms: H1 -- r, H0 -- r