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Proof step 118 from proof refine-open (previous step)

root[path].dir? ↔ ino0 ∈ dirs, fd = ino1, ofh = abs-of(of), root.dir?, dom of < ino2, abs(root, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(root, abs-files(files), abs-of(of)), ispath(path, ROOT_INO, ino0, dirs, files), of-cons(of, dirs, files), px-parent(user, path, ROOT_INO, dirs, files), ⌊err1⌋ , psynced → asynced, ino0 ∈ dirs ∨ ino0 ∈ files ⊦ ⟨err0 := err1; if err0 = ESUCCESS then let fid = root[path].fid in choose n with ¬ n ∈ ofh in {ofh[n] := mkfh(fid, mode, 0); fd := n}⟩ (ofh = abs-of(of) ∧ (psynced → asynced) ∧ ino1 = fd ∧ err1 = err0)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 119:
root[path].dir? ↔ ino0 ∈ dirs, err0 = err1, fd = ino1, ofh = abs-of(of), root.dir?, dom of < ino2, abs(root, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(root, abs-files(files), abs-of(of)), ispath(path, ROOT_INO, ino0, dirs, files), of-cons(of, dirs, files), px-parent(user, path, ROOT_INO, dirs, files), ⌊err1⌋ , psynced → asynced, ino0 ∈ dirs ∨ ino0 ∈ files ⊦ ⟨if err0 = ESUCCESS then let fid = root[path].fid in choose n with ¬ n ∈ ofh in {ofh[n] := mkfh(fid, mode, 0); fd := n}⟩ (ofh = abs-of(of) ∧ (psynced → asynced) ∧ ino1 = fd ∧ err1 = err0)

Proof step 118 from proof refine-open (previous step)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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