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Proof step 11 from proof replayone-keep-adr (previous step)

adr0 ∈ (inodekey(key.ino) ∈ ri ⊃ ri[key, adr];ri) ⊦ adr0 = adr, adr0 ∈ ri

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ adr ≠ adr0 ∧ ¬ adr ∈ ri → ¬ adr ∈ ri[key, adr0]; (s)