P is contained in NP

If you can SOLVE a problem quickly, you can certainly CHECK an answer to it quickly: just solve it yourself and compare. That one-line observation is the whole reason every problem in P is also in NP. Solving is at least as good as checking, so the efficiently-solvable problems sit inside the efficiently-checkable ones. Written with the subset symbol ⊆ (is contained in): P ⊆ NP.

Here is the proof in slow motion. Take any problem in P; by definition it has a deterministic algorithm A that runs in polynomial time and outputs the correct yes/no answer. To show it is in NP we must exhibit a polynomial-time verifier. Use this one: ignore the certificate entirely, run A on the input, and accept exactly when A says 'yes'. For 'yes' instances, ANY certificate (even the empty string) works, so a certificate exists; for 'no' instances, A says 'no' regardless, so none works. The verifier runs in polynomial time because A does. Hence the problem is in NP, and since the problem was arbitrary, all of P is in NP.

This containment is the firm, proven part of the P versus NP picture, and it sets the stage for the real question. We know P ⊆ NP for certain. What we do not know is whether the containment is strict, that is, whether NP holds problems that P does not. Almost everyone conjectures it is strict (P is not equal to NP), meaning there are problems easy to verify but genuinely hard to solve, but a proof has resisted decades of effort. Keep the logic straight: P ⊆ NP is a theorem; P is not equal to NP is a conjecture.