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Proof step 172 from proof inv-rmdir (previous step)

(entries[str] = ino1 ⊃ mkdir(m - 1, md1, # links(entries[str], dirs0), m2 - 1, entries -- str) ;mkdir(0, md, # links(entries[str], dirs0), m1, ∅) ) = mkdir(m0, md0, m3, m4, entries0), dirs0[ino1] = mkdir(m, md1, # links(ino1, dirs0), m2, entries), dirs0[entries[str]] = mkdir(0, md, # links(entries[str], dirs0), m1, ∅), dirs = dirs0[entries[str], mkdir(m0, md0, m3 - 1, m4, entries0)] [ino1, (ino1 = entries[str] ⊃ mkdir(m0, md0, m3 - 1, m4, entries0);mkdir(m - 1, md1, # links(ino1, dirs0), m2 - 1, entries -- str)).meta:= md1 ], files = files0[ino, files0[ino].size:= inode.size][ino, files0[ino, files0[ino].size:= inode.size][ino].meta:= inode.meta], ino = inode.ino, inode = mkinode(entries[str], md, true, # links(entries[str], dirs0), m1, 0), dom of < ino0, afs-cons(dirs0, files0), ispath(q, ROOT_INO, ino1, dirs0, files0), of-cons(of, dirs0, files0), pw(user, md1), px(user, md1), px-parent(user, q, ROOT_INO, dirs0, files0), entries[str] ∈ dirs0, str ∈ entries, ino1 ∈ dirs0, ino ∈ files0, 1 ≤ m3, 1 ≤ m2, 1 ≤ m, links(ino1, dirs0) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ [c_inode := getinode(entries[str], dirs, files); err := ESUCCESS; asynced := ?] (afs-cons(dirs, files) → ⟪vfs_put_inode#(c_inode, of, asynced; err, dirs, files)⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files))), entries[str] = ROOT_INO, m = 0, m2 = 0, m3 = 0

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mkinode(m, md, boolvar, m0, m1, m2).ino = m; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).size = m2; (s)% ⊦ files[ino, af][ino0] = (ino0 = ino ∧ ino ∈ files ⊃ af;files[ino0]); (s)% ⊦ afs-cons(dirs, files) ∧ ino ∈ dirs → ¬ ino ∈ files; (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).meta = md; (s)% ⊦ files[ino, af][ino, af0] = files[ino, af0]; (s)