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Proof step 198 from proof refine-read (previous step)

files[m3] = mkfile(m1, md, m2, content), of[fd] = mkfile(m3, mode0, m4), err0 = ESUCCESS, err0 = err1, fh = ofh[fd], 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)), of-cons(of, dirs, files), read-inv(m4, m4 + n0, mkinode(m3, md, false, m2, 0, m1), buf0, buf0, 0, dirs, files), read-inv(m4, m4 + n0, mkinode(m3, md, false, m2, 0, m1), buf0, buf1, 0, dirs, files), fd ∈ of, fd ∈ ofh, n0 ≤ # buf0, m4 ≤ m1, psynced → asynced, ⌊err1⌋ ∨ err1 = ESUCCESS ⊦ fh.fid ∈ abs-files(files), mode0 ≠ MODE_R ∧ mode0 ≠ MODE_RW

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)% ⊦ ¬ ⌊ESUCCESS⌋ ; (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)