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Proof step 133 from proof recovery (previous step)

fs0[(ri1 ⊖ orphans(fs0, ri1))[inodekey(ino)]].nlink = 0, orphans(fs0, ri1).inodes?, fo0.inodes?, afs-cons(abs-dirs(fs0, ri1), abs-files(fs0, ri1)), afs-cons(abs-dirs(fs0, ri1) \ abs-keys(orphans(fs0, ri1)), abs-files(fs0, ri1) \ abs-keys(orphans(fs0, ri1))), file-node-cons(fs0, ri1), file-node-cons(fs0, ri1 ⊖ orphans(fs0, ri1)), log-cons(log0, fs0), orphans-cons(fs0, ri1, orphans(fs0, ri1)), replay-cons-strong(fs0, ri1 ⊖ orphans(fs0, ri1), ∅, fi0, fo0, log0), replay-safe(log0, fs0, fi0 ⊖ fo0), ri-inj(ri1), ri-inj(fi0), ri-inj(ri1 ⊖ orphans(fs0, ri1)), ubifs-cons(fs0, ri1, orphans(fs0, ri1), fi0, fo0, log0), ubifs-cons(fs0, ri1 ⊖ orphans(fs0, ri1), ∅, fi0, fo0, log0), inodekey(ino) ∈ ri1, ∀ ino0. inodekey(ino0) ∈ ri1 ⊖ orphans(fs0, ri1) → fs0[(ri1 ⊖ orphans(fs0, ri1))[inodekey(ino0)]].nlink ≠ 0 ∨ ino0 = ROOT_INO ⊦ # abs-keys(orphans(fs0, ri1 ⊖ orphans(fs0, ri1))) = 0, fs0[ri1[inodekey(ino)]].nlink = 0, ino = ROOT_INO

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 18 terms: ino

Additional information

Used simplifier rules: ⊦ inodekey(ino) ∈ ri → (inodekey(ino) ∈ ri ⊖ ro ↔ ¬ inodekey(ino) ∈ ro); (s)%⊦ key ∈ orphans(fs, ri) ↔ key.inode? ∧ key ∈ ri ∧ fs[ri[key]].nlink = 0 ∧ key.ino ≠ ROOT_INO; (s)% ⊦ inodekey(m).inode?; (s) Used terms: ino