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Proof step 51 from proof inv-create (previous step)

⟨err := ESUCCESS⟩ ( mkdentry(dent.name, ino) = dent0 ∧ err = err0 ∧ dirs[p_ino, dirs[p_ino].entries:= dirs[p_ino].entries[dent.name, ino]] [p_ino, dirs[p_ino, dirs[p_ino].entries:= dirs[p_ino].entries[dent.name, ino]][p_ino] .size:= (dirs[p_ino, dirs[p_ino].entries:= dirs[p_ino].entries[dent.name, ino]][p_ino].size + 1) ] = dirs0 ∧ (files ++ ino)[ino, mkfile(0, md, 1, ∅)] = files0), dirs1[m] = mkdir(m0, md2, # links(m, dirs1), m2, entries0), dent = negdentry(str), dirs = dirs1, files = files1, ino = ino0, p_ino = m, afs-cons(dirs0, files0), afs-cons(dirs1, files1), afs-cons(dirs0[m, dirs0[m].meta:= md0], files0), m ∈ dirs0, m ∈ dirs1, links(m, dirs1) ≠ ∅ ∨ m = ROOT_INO ⊦ ino0 = 0, dent0.dentry?, ino0 ∈ dirs1, ino0 ∈ files1, str ∈ entries0

Heuristic symbolic execution applied rule assign left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ negdentry(str).name = str; (s)% ⊦ dirs[ino, ad][ino0] = (ino0 = ino ∧ ino ∈ dirs ⊃ ad;dirs[ino0]); (s)% ⊦ (ad.entries:= entries).size = ad.size; (s)% ⊦ dirs[ino, ad][ino, ad0] = dirs[ino, ad0]; (s)% ⊦ ino ∈ dirs[ino0, ad] ↔ ino ∈ dirs; (s)% ⊦ mkdentry(str, m).dentry?; (s)