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

ro.inodes?, ∀ ino. ino ∈ abs-keys(ro) ↔ ino ∈ abs-keys(orphans(fs, ri)) ⊦ ro = orphans(fs, ri)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ro.inodes?, ∀ ino. inodekey(ino) ∈ ro ↔ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].nlink = 0 ∧ ino ≠ ROOT_INO ⊦ ro = orphans(fs, ri)

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)