Question

Question: Show that if P=NP, then P=PH

Short answer

Reduce ${\mathbf{II}}_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{P}$machine to$\mathbf{P}\mathbf{=}{\mathbf{C}}_{\mathbf{o}}\mathbf{NP}$ machine, hence, putting it in P as well, and completing collapse of hierarchy.

Step-by-step solution

Step 1:  Assuming ∑i+1 machine M

• Firstly, if $\mathbf{P}\mathbf{=}\mathbf{NP}$, then because P is closed under complement, thus $\mathbf{P}\mathbf{=}{\mathbf{C}}_{\mathbf{O}}\mathbf{NP}$. Written as, $\mathbf{P}\mathbf{=}\mathbf{\sum }_{\mathbf{i}}\mathbf{P}\mathbf{=}{\mathbf{II}}_{\mathbf{i}}\mathbf{P}$.
• Now using induction that if $\mathbf{P}\mathbf{=}\mathbf{\sum }_{\mathbf{i}}\mathbf{P}\mathbf{=}{\mathbf{II}}_{\mathbf{i}}\mathbf{P}$then $\mathbf{P}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{P}\mathbf{=}{\mathbf{II}}_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{P}$
• Assume$\mathbf{\sum }_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{p}$ machine M, that consists of a run of the existential branching, then existential branching etc.
• Assume the computation sub-tree path whose root are first universal step along path. For each such type of sub-tree, M is performing a computation. By hypothesis, ${\mathbf{II}}_{\mathbf{i}}\mathbf{}\mathbf{P}\mathbf{=}\mathbf{P}$.
• Thus, for the forming of new machine$\mathit{S}$ each of computation sub-trees can be replaced by deterministic (non- branching) polynomial time of computation.

Step 2: Assuming  a(n) be the number of maximum steps which are taken by other machine before the start of universal machine

• If assume a(n) be the number of maximum steps which are taken by other machine before the start of universal machine, P(n) be the maximum steps which are taken by any deterministic which were substituted for computations in P machines, therefore S is covered by $\mathbf{a}\mathbf{\left(}\mathbf{n}\mathbf{\right)}\mathbf{+}\mathbf{p}\mathbf{\left(}\mathbf{a}\mathbf{\right(}\mathbf{n}\mathbf{\left)}\mathbf{\right)}$.
• Remember that the $\mathbf{p}\mathbf{\left(}\mathbf{a}\mathbf{\right(}\mathbf{n}\mathbf{\left)}\mathbf{\right)}$ term is composition of the functions, because P sub procedures with inputs are computing which may be a longer than (but it must be equal or smaller than a(n), since only a(n) steps are executed on the time the sub procedures are used).
• Since aand pboth are polynomials, therefore S, is in NP. By hypothesis, $\mathbf{P}\mathbf{=}\mathbf{NP}$so S is in P as well.

A similar type of argument may be used to reduce a${\mathbf{II}}_{\mathbf{i}\mathbf{+}\mathbf{1}}\mathbf{P}$ machine to$\mathbf{P}\mathbf{=}{\mathbf{C}}_{\mathbf{o}}\mathbf{NP}$ machine, hence, putting it in P as well, and completing collapse of hierarchy.