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Proof step 179 from proof refine-mkdir (previous step)

dirs0[ino1] = mkdir(# entries0, md1, # links(ino1, dirs0), m1, entries0), err0 = ESUCCESS, ((ino2 => mkdir(0, md, 1, 0, ∅) * (λ dirs1. dirs3 = dirs1)))(dirs1), root.dir?, root[q].dir?, dom of < ino0, abs(root, ROOT_INO)(dirs0), afs-cons(dirs0, files), afs-cons(dirs, files), fids-cons(root, abs-files(files), abs-of(of)), ispath(q, ROOT_INO, ino1, dirs0, files), of-cons(of, dirs0, files), of-cons(of, dirs, files), pw(user, md1), px(user, md1), px-parent(user, q, ROOT_INO, dirs0, files), ino2 ∈ dirs4, ino1 ∈ dirs0, ino1 ∈ dirs1, dirs1 ≡ dirs4[ino2, mkdir(0, md, 1, 0, ∅)], dirs4 ≡ dirs3 ++ ino2, dirs3 ≡ dirs0[ino1, mkdir(# entries0 + 1, md1, # links(ino1, dirs0), m1 + 1, entries0[str, ino2])], dirs ≡ dirs1[ino1, dirs1[ino1].meta:= md1], psynced → asynced0, links(ino1, dirs0) ≠ ∅ ∨ ino1 = ROOT_INO ⊦ ⟨psynced := ?⟩ (abs(root[q + str ', dnode(md, ∅)], ROOT_INO)(dirs) ∧ (psynced → asynced) ∧ ESUCCESS = err0), ino2 ∈ dirs0, ino2 ∈ files, str ∈ entries0, ino2 ∈ dirs3

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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