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Proof step 44 from proof refine-write (previous step)

files0 -- m2 = files -- m2, (content.stream to m).buf = splice(buf, 0, (content0.stream to m1).buf, m3, n1), of[fd] = mkfile(m2, mode, m3), files0[m2] = mkfile(m, md, m0, content), files[m2] = mkfile(m1, md, # links(m2, dirs), content0), err0 = ESUCCESS, fh = ofh[fd], fs = abs-files(files), ofh = abs-of(of), root.dir?, dom of < ino, abs(root, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(root, abs-files(files), abs-of(of)), of-cons(of, dirs, files), fd ∈ of, m2 ∈ files, m2 ∈ files0, fd ∈ ofh, n1 ≤ # buf, n0 ≤ # buf, n1 ≤ n0, psynced → asynced, mode = MODE_W ∨ mode = MODE_RW ⊦ fh.fid ∈ fs

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ of-cons(of, dirs, files) ∧ fd ∈ of ∧ of[fd] = mkfile(ino, mode, n) → ino ∈ files; (ss)% ⊦ mkfile(m, mode, m0).ino = m; (s)% ⊦ ⌜ino⌝ ∈ abs-files(files) ↔ ino ∈ files; (s)% ⊦ mkfh(fid, mode, m).fid = fid; (s)%fd ∈ of ⊦ abs-of(of)[fd] = mkfh(⌜of[fd].ino⌝ , of[fd].mode, of[fd].pos); (s)% ⊦ fd ∈ abs-of(of) ↔ fd ∈ of; (s)