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Proof step 2 from proof at-del-ri (previous step)

⊦ (ri -- key0)[key1] = (key0 = key1 ⊃ (∅)[key1];ri[key1])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ key ≠ key0 → (ri -- key0)[key] = ri[key]; (s)% ⊦ (ri -- key)[key] = (∅)[key]; (s)