a running turing machine will eventually do exactly one of the following three things: stop and accept, stop and reject or so called loop (i.e. never stop). we proved in class with a diagonalization technique that there can be no turing machine h which when given as input the encoding for an arbitrary turing machine, m, and an arbitrary input string, x, can decide if m will stop, (i.e., accept or reject), on input x or loop. the proof is a proof by contradiction. what is the contradiction that is arrived at in the proof?