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