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Proof step 22 from proof initialization (previous step)

⊦ ∃ root, fs, ofh. ⟨choose err with ⌊err⌋ in err0 := err⟩ (ofh = ∅ ∧ fs = ∅ ∧ abs(root, ROOT_INO)((∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)]) ∧ ESUCCESS = err0), ∃ ino, ino0. ((ROOT_INO => mkdir(0, md, ino, ino0, ∅)))((∅ ++ ROOT_INO)[ROOT_INO, mkdir(0, md, 0, 0, ∅)])

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: 0, 0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ((ino => ad))(dirs) ↔ ino ≠ 0 ∧ dirs = (∅ ++ ino)[ino, ad]; (ss)% ⊦ ROOT_INO ≠ 0; (s) Used terms: 0, 0