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Proof step 6 from proof inv-read (previous step)

err0 = err1, n0 = n, root.dir?, fids-cons(root, fs, ofh), pre-read(fd, n0, err, user, root, fs, ofh), n ≤ # buf ⊦ ⟪if err = ESUCCESS then let fh = ofh[fd] in let f = fs[fh.fid] in choose n with n ≤ n0 ∧ fh.pos + n ≤ # f.content in {n0 := n; buf := copy(f.content, fh.pos, buf, 0, n0); ofh[fd].pos := fh.pos + n0} ifnone n0 := 0 else n0 := ?⟫ fids-cons(root, fs, ofh)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Proof step 7:
err = ESUCCESS, err0 = err1, n0 = n, root.dir?, fids-cons(root, fs, ofh), pre-read(fd, n0, err, user, root, fs, ofh), n ≤ # buf ⊦ ⟪let fh = ofh[fd] in let f = fs[fh.fid] in choose n with n ≤ n0 ∧ fh.pos + n ≤ # f.content in {n0 := n; buf := copy(f.content, fh.pos, buf, 0, n0); ofh[fd].pos := fh.pos + n0} ifnone n0 := 0⟫ fids-cons(root, fs, ofh)
Proof step 21:
err0 = err1, n0 = n, root.dir?, fids-cons(root, fs, ofh), pre-read(fd, n0, err, user, root, fs, ofh), n ≤ # buf ⊦ ⟪n0 := ?⟫ fids-cons(root, fs, ofh), err = ESUCCESS

Rule argument

pos: right 1

Proof step 6 from proof inv-read (previous step)

Heuristic conditional right split applied rule if right

And obtained the following new goals

Rule argument

Additional information