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Proof step 10 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) = ∅, ¬ ino ∈ dirs ∧ ¬ ino ∈ files

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # ls ≠ 0 → ls ≠ ∅; (forward) from specification set-basic[alink-set]%size-zero-01: ⊦ ls ≠ ∅ → # ls ≠ 0; (forward) from specification set-basic[alink-set]