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Proof step 12 from proof afs-cons-mkdir-ino (previous step)

dirs[m] = mkdir(m0, md0, # links(m, dirs), m2, entries), dent = negdentry(str), p_ino = m, afs-cons(dirs, files), m ∈ dirs, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ ⟪dirs := 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) ] [p_ino, 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) ] [p_ino ] .nsubdirs:= ( 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) ] [p_ino ] .nsubdirs + 1) ]; dirs := dirs ++ ino; dirs[ino] := mkdir(0, md, 1, 0, ∅); dent := mkdentry(dent.name, ino); err := ESUCCESS⟫ afs-cons(dirs, files), ino = 0, ino ∈ files, ino ∈ dirs, str ∈ entries

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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