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Proof step 10 from proof abs-keys-orphans2 (previous step)

∀ ino. ino ∈ abs-keys(orphans(fs0, ri0)) ↔ ino ∈ abs-keys(orphans(fs, ri)) ⊦ inodekey(m) ∈ ri0 ∧ fs0[ri0[inodekey(m)]].nlink = 0 ↔ inodekey(m) ∈ ri ∧ fs[ri[inodekey(m)]].nlink = 0, m = ROOT_INO

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ ino. ino ≠ ROOT_INO → (inodekey(ino) ∈ ri0 ∧ fs0[ri0[inodekey(ino)]].nlink = 0 ↔ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].nlink = 0) ⊦ inodekey(m) ∈ ri0 ∧ fs0[ri0[inodekey(m)]].nlink = 0 ↔ inodekey(m) ∈ ri ∧ fs[ri[inodekey(m)]].nlink = 0, m = ROOT_INO

Rule argument

Additional information

Used simplifier rules: abs-keys: ⊦ n ∈ abs-keys(ks) ↔ inodekey(n) ∈ ks;% ⊦ inodekey(m).ino = m; (s)% ⊦ inodekey(m).inode?; (s)%⊦ key ∈ orphans(fs, ri) ↔ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO; (s)