:load break026
:step foldl (+) 0 [1..5]
:step
:step
:step
:step
:step
:force c
        -- answer should be 1

:load break026
:step foldl (+) 0 [1..5]
:step
:step
:step
:step
:step
-- a diversion to single-step the evaluation of c:
:step c `seq` ()
:step
-- end diversion
c
        -- answer should be 1 again (not 0)