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Proof step 14 from proof repalyone-monotonic (previous step)

dentrynode(key0, ino) = fs[adr], key ∈ (ino ≠ 0 ⊃ ri0[key0, adr];ri0 -- key0), key ∈ ri0 → key ∈ ri1, ∀ key. key ∈ ri0 → key ∈ ri1 ⊦ key ∈ (ino ≠ 0 ⊃ ri1[key0, adr];ri1 -- 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 ∈ ri -- key0 ↔ key ≠ key0 ∧ key ∈ ri; (s)% ⊦ key ∈ ri[key0, adr] ↔ key = key0 ∨ key ∈ ri; (ws)% ⊦ key ≠ key0 → (key ∈ ri[key0, adr] ↔ key ∈ ri); (s)