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Proof step 14 from proof disjoint-union-empty (previous step)

a0 ∈ st1, ∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
a0 ∈ st1, ∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦

Rule argument

Additional information