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Proof step 1 from proof at-ri (conclusion)

⊦ ri[key, adr][key0] = (key = key0 ⊃ adr;ri[key0])

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, adr][key] = ri[key]; (s)% ⊦ ri[key, adr][key] = adr; (s)