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Proof step 5 from proof replayone-keep-adr (previous step)

adr0 ∈ (ino ≠ 0 ∨ key.ino = ROOT_INO ⊃ ri[key, adr];(boolvar ⊃ ri -- key;ri ⊖ key)) ⊦ adr0 = adr, adr0 ∈ ri

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
ino = 0, adr0 ∈ (boolvar ⊃ ri -- key;ri ⊖ key) ⊦ key.ino = ROOT_INO, adr0 = adr, adr0 ∈ ri

Rule argument

pos: left 1

Additional information

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