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Proof step 99 from proof refine-seek (previous step)

n = n0, ofh = abs-of(of), whence ≠ SEEK_END, whence ≠ SEEK_CUR, 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, psynced → asynced ⊦ fd ∈ ofh ∧ ⟨err0 := ESUCCESS; if err0 = ESUCCESS then let fh = ofh[fd] in let f = abs-files(files)[fh.fid] in {if whence = SEEK_CUR then n := fh.pos + n else if whence = SEEK_END then n := # f.content + n; ofh[fd] := mkfh(ofh[fd].fid, ofh[fd].mode, n)}⟩ (ofh = abs-of(of[fd, of[fd].pos:= n]) ∧ (psynced → asynced) ∧ n0 = n ∧ ESUCCESS = err0)

Heuristic dl case distinction applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ fd ∈ abs-of(of) ↔ fd ∈ of; (s)