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Proof step 10 from proof afs-cons-reduce (previous step)

files0[ino] = mkfile(m, md, m0, content), files1[ino] = mkfile(m1, md0, m2, content0), dirs0[ino] = mkdir(m3, md1, m4, m5, entries), afs-cons(dirs0, files0), afs-cons(dirs1, files1), ROOT_INO ∈ dirs0, ino ∈ dirs0 ∧ name ∈ entries → entries[name] ∈ dirs0 ∨ entries[name] ∈ files0, ino ∈ files0 → ¬ ino ∈ dirs0 ∧ m0 = # links(ino, dirs0), ino ∈ dirs0 → m4 = # links(ino, dirs0) ∧ m4 ≤ 1 ∧ m3 = # entries ∧ m5 = # subdirs(ino, dirs0), ino ∈ dirs0 ∧ ino ≠ ROOT_INO ∧ links(ino, dirs0) = ∅ → entries = ∅ ⊦ (ino ∈ dirs1 ∧ ino ≠ ROOT_INO ∧ links(ino, dirs1) = ∅ → dirs1[ino].entries = ∅) ∧ ( ino ∈ dirs1 → dirs1[ino].nlink = # links(ino, dirs1) ∧ dirs1[ino].nlink ≤ 1 ∧ dirs1[ino].size = # dirs1[ino].entries ∧ dirs1[ino].nsubdirs = # subdirs(ino, dirs1)) ∧ (ino ∈ files1 → ¬ ino ∈ dirs1 ∧ m2 = # links(ino, dirs1)) ∧ (ino ∈ dirs1 ∧ name ∈ dirs1[ino].entries → dirs1[ino].entries[name] ∈ dirs1 ∨ dirs1[ino].entries[name] ∈ files1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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