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Proof step 35 from proof refine-seek (previous step)

of[fd] = mkfile(m, mode, m0), err0 = ESUCCESS, f = abs-files(files)[fh.fid], fh = ofh[fd], n = n0, 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 ∈ ofh, fd ∈ of, fh.fid ∈ abs-files(files), psynced → asynced ⊦ ⟨ofh[fd] := mkfh(ofh[fd].fid, ofh[fd].mode, fh.pos + n)⟩ (ofh = abs-of(of[fd, of[fd].pos:= (m0 + n0)]) ∧ (psynced → asynced) ∧ m0 + n0 = fh.pos + n ∧ ESUCCESS = err0)

Heuristic symbolic execution applied rule assign right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ fd ∈ abs-of(of) ↔ fd ∈ of; (s)%fd ∈ of ⊦ abs-of(of)[fd] = mkfh(⌜of[fd].ino⌝ , of[fd].mode, of[fd].pos); (s)% ⊦ mkfh(fid, mode, m).fid = fid; (s)% ⊦ ⌜ino⌝ ∈ abs-files(files) ↔ ino ∈ files; (s)% ⊦ mkfh(fid, mode, m).mode = mode; (s)% ⊦ mkfh(fid, mode, m).pos = m; (s)%⊦ abs-of(of)[fd, mkfh(⌜ino⌝ , mode, n)] = abs-of(of[fd, mkfile(ino, mode, n)]); (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ mkfile(m, mode, m0).ino = m; (s)% ⊦ mkfile(m, mode, m0).mode = mode; (s)% ⊦ mkfile(m, mode, m0).pos = m0; (s)% ⊦ mkfile(m, mode, m0).pos:= m2 = mkfile(m, mode, m2); (s)