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Proof step 15 from proof abs-keys-orphans-add2 (previous step)

∀ 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 ⊦ inodekey(ino) ∈ ri0 ∧ fs0[ri0[inodekey(ino)]].nlink = 0 ∧ ino ≠ ROOT_INO ↔ ino = n ∨ inodekey(ino) ∈ ri ∧ fs[ri[inodekey(ino)]].nlink = 0 ∧ ino ≠ ROOT_INO

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: inodekey(ino) abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ inodekey(m).inode?; (s)% ⊦ inodekey(m).ino = m; (s)% ⊦ inodekey(m) = inodekey(m0) ↔ m = m0; (s) Used terms: inodekey(ino)