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Proof step 6 from proof new-exists (previous step)

∀ n. n ∈ dirs ∨ n ∈ files → n < ino ⊦ ino = 0, ∃ ino. ino ≠ 0 ∧ ¬ ino ∈ dirs ∧ ¬ ino ∈ files, ¬ ino ∈ dirs ∧ ¬ ino ∈ files

User applied rule case distinction

And obtained the following new goals

Proof step 7:
ino ∈ dirs, ∀ n. n ∈ dirs ∨ n ∈ files → n < ino ⊦ ino = 0, ∃ ino. ino ≠ 0 ∧ ¬ ino ∈ dirs ∧ ¬ ino ∈ files
Proof step 9:
ino ∈ files, ∀ n. n ∈ dirs ∨ n ∈ files → n < ino ⊦ ino = 0, ∃ ino. ino ≠ 0 ∧ ¬ ino ∈ dirs ∧ ¬ ino ∈ files

Rule argument

pos: right 3

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