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Proof step 1350 from proof inv-renam (previous step)

dirs[ino1] = mkdir(m0, md0, # links(ino1, dirs), m3, entries0), dirs[ino] = mkdir(m, md, # links(ino, dirs), m2, entries), err = EACCESS, err0 = ENOENT, err1 = ESUCCESS, inode = getinode(ino, dirs, files), inode0 = mkinode(ino, md, true, # links(ino, dirs), m2, m), mkdentry(str, entries0[str]).dentry?, dom of < ino2, afs-cons(dirs, files), ispath(q, ROOT_INO, ino1, dirs, files), ispath(q0, ROOT_INO, ino, dirs, files), of-cons(of, dirs, files), pw(user, md0), px(user, md), px(user, md0), px-parent(user, q, ROOT_INO, dirs, files), px-parent(user, q0, ROOT_INO, dirs, files), str ∈ entries0, ino ∈ dirs, entries0[str] ∈ files, ino1 ∈ dirs, links(ino1, dirs) ≠ ∅ ∨ ino1 = ROOT_INO, links(ino, dirs) ≠ ∅ ∨ ino = ROOT_INO, ¬ px(user, inode.meta) ∨ ¬ pw(user, inode.meta) ⊦ ⟪skip⟫ (of-cons(of, dirs, files) ∧ afs-cons(dirs, files)), str0 ∈ entries, 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: ⊦ mkdentry(str, m).dentry?; (s)%⊦ 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)