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Proof step 11 from proof disjoint-union-empty (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 12:
a ∈ st0, ∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦
Proof step 14:
a0 ∈ st1, ∀ a. ¬ a ∈ st0 ∪ st1 ∧ st1[a] = (∅)[a] ⊦

Rule argument

pos: right 1

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