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Proof step 8 from proof file-node-cons-variant1-01 (previous step)

inodekey(ino) ∈ ri0, inodekey(ino) ∈ ri ∧ ¬ fs[ri[inodekey(ino)]].directory → fs[ri[inodekey(ino)]].nsubdirs = 0, ∀ ino. inodekey(ino) ∈ ri0 ∧ ¬ fs0[ri0[inodekey(ino)]].directory → fs0[ri0[inodekey(ino)]].nsubdirs = 0, ∀ ino. inodekey(ino) ∈ ri ∧ ¬ fs[ri[inodekey(ino)]].directory → fs[ri[inodekey(ino)]].nsubdirs = 0 ⊦ fs0[ri0[inodekey(ino)]].nsubdirs = 0, fs0[ri0[inodekey(ino)]].directory

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: ino

Additional information

Used terms: ino