Batch: Shutdown PC if a Program isn't Running?

Question

Is there a batch command that will check if a program is running and shutdown the PC if it isn't?

Answer 1

Thats Possible, example looking for running Internet explorer:

@echo off

tasklist /fi "imagename eq iexplore.exe" | find /I "iexplore.exe" > nul
if errorlevel 1 goto SHUTDOWN

echo IE is already running
goto DONE

:SHUTDOWN
shutdown -s -f -t 0

goto DONE

:DONE
exit

EDIT

if you want to Loop this command use this:

@echo off
:loop
tasklist /fi "imagename eq iexplore.exe" | find /I "iexplore.exe" > nul
if errorlevel 1 goto SHUTDOWN

echo IE is already running
goto DONE

:SHUTDOWN
shutdown -s -f -t 0

goto DONE

:DONE
REM wait 5 sec
ping 127.0.0.1 -n 2 -w 5000 > NUL
goto loop

Answer 2

Just for a one liner

for /l %%# in (1 0 2) do (tasklist|find /i "program.exe">nul ||(shutdown -s -f -t 0 & exit /b))

Create a loop starting in 1 until 2 with a step of 0, check if program.exe in tasklist output. If it fails run shutdown and exit.