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Proof step 4 from proof path-links-empty (previous step)

p = ε, ispath(p, ino, n, dirs, files) ⊦ # links(n, dirs) = 0 ↔ p = ε ∧ links(ino, dirs) = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
ino ∈ dirs ∨ ino ∈ files ⊦ # links(ino, dirs) = 0 ↔ links(ino, dirs) = ∅

Rule argument

Additional information

Used simplifier rules: path-here: ⊦ ispath(ε, n, m, dirs, files) ↔ valid-ino(n, dirs, files) ∧ m = n;%valid-ino: ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files;