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Proof step 5 from proof abs-keys-orphans-add (previous step)

abs-keys(ro) = abs-keys(orphans(fs, ri)) ++ n, ro.inodes? ⊦ ∀ key. key ∈ ro ↔ key ∈ orphans(fs, ri) ++ inodekey(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs-keys(ro) = abs-keys(orphans(fs, ri)) ++ n, ro.inodes? ⊦ key ∈ ro ↔ key = inodekey(n) ∨ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO

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)% ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)