Final answer:
The problem GE5 is undecidable by using a reduction from the Halting Problem. By assuming a decider for GE5, we can construct a machine that uses the supposed decider to solve the Halting Problem, leading to a contradiction since the Halting Problem is undecidable.
Step-by-step explanation:
The question asks to show that the problem GE5, which is defined as the set containing an encoding of an algorithm M such that the language accepted by M, denoted L(M), has at least five elements, is undecidable. To prove that GE5 is undecidable, we can use a reduction from the Halting Problem. The Halting Problem is known to be undecidable, and it asks whether a given Turing machine M halts for an input w.
We can construct a reduction to prove that GE5 is undecidable by assuming that there is a decider D for GE5, and then use this supposed decider to solve the Halting Problem, which leads to a contradiction. Specifically, we can construct a Turing machine M' that, on an input, simulates M on w and includes additional code to produce at least five different outputs if M halts. If M' enters an infinite loop, it produces fewer than five outputs. A decider D for GE5 would then be able to determine whether M halts on w by checking if M' has at least five elements in its language, effectively solving the Halting Problem. This contradiction implies that GE5 cannot have a decider, hence it is undecidable.