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Proof step 12 from proof orphans-cons-gc-one (previous step)

ri-cons(ri, fs), inodekey(ino) ∈ ri, ∀ ino. inodekey(ino) ∈ ri → (fs[ri[inodekey(ino)]].nlink = 0 ∧ ino ≠ ROOT_INO ↔ inodekey(ino) ∈ ro) ⊦ fs[ri[inodekey(ino)]].nlink = 0 ∧ ino ≠ ROOT_INO ↔ inodekey(ino) ∈ ro, adr ∈ fs

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ino

Additional information

Used terms: ino