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Proof step 1 from proof eq-union-disj (conclusion)

⊦ zh0 ⊥ zh ∧ zh1 ⊥ zh → (zh0 ∪ zh = zh1 ∪ zh ↔ zh0 = zh1)

System applied rule normalize

And obtained the following new goals

Proof step 2:
zh0 ⊥ zh, zh1 ⊥ zh ⊦ zh0 ∪ zh = zh1 ∪ zh ↔ zh0 = zh1

Rule argument

Additional information