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Proof step 1 from proof at-diff-ri (conclusion)

⊦ (ri \ ks)[key] = (key ∈ ks ⊃ (∅)[key];ri[key])

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ key ∈ ks → (ri \ ks)[key] = ri[key]; (s)% ⊦ key ∈ ks → (ri \ ks)[key] = (∅)[key]; (s)