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Proof step 965 from proof inv-renam (previous step)

isdir0 ↔ isdir1, isdir1 ↔ false, isdir ↔ false, files[entries0[str]] = mkfile(m1, md1, # links(entries0[str], dirs), content), dirs[ino] = mkdir(m, md, # links(ino, dirs), m2, entries), dirs[ino1] = mkdir(m0, md0, # links(ino1, dirs), m3, entries0), dent = mkdentry(str0, entries[str0]), err = EACCESS, err0 = ESUCCESS, err1 = ESUCCESS, inode = getinode(ino, dirs, files), inode0 = mkinode(ino, md, true, # links(ino, dirs), m2, m), p_inode = mkinode(entries0[str], md1, false, # links(entries0[str], dirs), 0, m1), dom of < ino2, afs-cons(dirs, files), ispath(q0, ROOT_INO, ino, dirs, files), ispath(q, ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pw(user, md0), px(user, md0), px(user, md), px-parent(user, q0, ROOT_INO, dirs, files), px-parent(user, q, ROOT_INO, dirs, files), str0 ∈ entries, ino1 ∈ dirs, entries0[str] ∈ files, ino ∈ dirs, str ∈ entries0, links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO, ¬ px(user, inode.meta) ∨ ¬ pw(user, inode.meta) ⊦ ⟪skip⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files)), entries0[str] ∈ dirs, q + str ' ⊑ q0 + str0 '

Heuristic symbolic execution applied rule skip right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ getinode(ino, dirs, files) = (ino ∈ dirs ⊃ mkinode (ino, dirs[ino].meta, true, dirs[ino].nlink, dirs[ino].nsubdirs, dirs[ino].size) ;mkinode (ino, files[ino].meta, false, files[ino].nlink, 0, files[ino].size) ); (s)% ⊦ mkinode(m, md, boolvar, m0, m1, m2).meta = md; (s)