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Proof step 4 from proof union-left-empty (previous step)

⊦ (a ∈ st → a ∈ st ∪ ∅) ∧ (a ∈ st ⊃ st[a];(∅)[a]) = st[a]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
⊦ (a ∈ st ⊃ st[a];(∅)[a]) = st[a]

Rule argument

pos: right 1

Additional information

Used simplifier rules: union-in: ⊦ a ∈ st0 ∪ st1 ↔ a ∈ st0 ∨ a ∈ st1;