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Proof step 2 from proof eq-union-disj (previous step)

zh0 ⊥ zh, zh1 ⊥ zh ⊦ zh0 ∪ zh = zh1 ∪ zh ↔ zh0 = zh1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
zh0 ∪ zh = zh1 ∪ zh, zh1 ⊥ zh, zh0 ⊥ zh ⊦ zh0 = zh1

Rule argument

Additional information