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Proof step 11 from proof afs-cons-links-empty (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: afs-cons-notfile: ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files;%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;