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Proof step 18 from proof repalyone-monotonic (previous step)

datanode(key0, buf) = fs[adr], inodekey(key0.ino) ∈ ri1, key ∈ (inodekey(key0.ino) ∈ ri0 ⊃ ri0[key0, adr];ri0), ∀ key. key ∈ ri0 → key ∈ ri1 ⊦ key = key0, key ∈ ri1, key ∈ ri0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: ⊦ key ≠ key0 → (key ∈ ri[key0, adr] ↔ key ∈ ri); (s)