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Proof step 5 from proof abs-keys-orphans2 (previous step)

abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ⊦ ∀ key. key ∈ orphans(fs0, ri0) ↔ key ∈ orphans(fs, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)), key.inode? ⊦ key.ino = ROOT_INO, key ∈ ri0 ∧ fs0[ri0[key]].nlink = 0 ↔ key ∈ ri ∧ fs[ri[key]].nlink = 0

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ orphans(fs, ri) ↔ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO; (s)