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Proof step 3 from proof abs-keys-orphans-add2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ++ n ↔ (∀ key. key.inode? ∧ key ∈ ri0 ∧ fs0[ri0[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO ↔ key = inodekey(n) ∨ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO)

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)%⊦ key ∈ orphans(fs, ri) ↔ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO; (s)