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Proof step 9 from proof afs-cons-links-empty (previous step)

dirs[ino] = mkdir(m1, md0, m2, m3, entries), files[ino] = mkfile(m, md, m0, content), links(ino, dirs) = ∅, afs-cons(dirs, files) ⊦ ¬ ino ∈ dirs ∧ ¬ ino ∈ files, (ino ∈ dirs → m2 = 0) ∧ (ino ∈ files → m0 = 0)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: afs-cons-dir-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs ∧ dirs[ino] = mkdir(n0, md, n, n1, entries) → # links(ino, dirs) = n;%afs-cons-file-nlink: ⊦ afs-cons(dirs, files) ∧ ino ∈ files ∧ files[ino] = mkfile(m, md, n, content) → # links(ino, dirs) = n;%size-zero: ⊦ # ls ≠ 0 → ls ≠ ∅; (forward) from specification set-basic[alink-set]