White sunlight is a mixture of a huge range of frequencies (i.e. colours), of which most - but not all - are in the range that are visible to the eye. White is not one single colour, it is a mix of many colours.
The eye has 3 different colour receptors (cone cells). Each type is most sensitive to a certain colour range. Their peak wavelengths are in the red, green and blue regions of the visual spectrum, respectively. The brain then mixes the signals from all three types, to produce the sensation of a "colour".
Because our eyes evolved to see in sunlight, if the red, blue and green levels of the perceived spectrum are approximately those of sunlight, we see that colour as being the same as that of sunlight - something we call "white". Again, white is a mix of colours.
There are other perceived colours that do not appear in the spectrum. A common example is brown. Like white, brown is a mixture of pure colours. No part of the electromagnetic spectrum is brown, but our brain interprets certain mixtures of wavelengths that way. Tree trunks reflect only those parts of the sunlight that make the brain think "brown".
In the same way, if something reflects all wavelengths of sunlight, the brain thinks "white". In artificial lighting, if we mix the appropriate amounts of red, blue and green light, all three types of cone cells get the right stimulation to make the brain perceive it like sunlight, i.e. white. If the mixture is not exactly right, the brain will see it as some other shade, like pink or grey.
In fact, different "white" lights have different colour shades. It is said they have a different "colour temperature". The temperature of the sun is close to 6000K and light of that colour temperature looks like sunlight. Many white LEDs have colour temperatures of around 9000K, and their light will have a bluish tinge. Incandescents globes are closer to 4000K and look reddish - but we call all of these lightsources "white".
Best Answer
Yes, but not with equal amounts of each.
In order to answer this, we need to understand the CIE 1931 color space, and think about its algebraic properties.
Essentially what the CIE specification says is that, while light comes to us as a spectrum filled with varying amounts of photons in the wavelength range 380-700nm, our eyes are engineered in such a way that they only have 3 receptors.
These rod and cone receptors act linearly on the frequency/wavelength distribution, and can be represented as 3 integrals of "color matching functions" against the distribution (linear functionals). One is sensitive mostly in the "red" region ($\approx$ 500-700nm), one in the "green" ($\approx$ 440-660nm, more spread, smaller peak), and one in the "blue" ($\approx$ 380-500nm).
The color matching functions are determined empirically, meaning by experiment. They got a bunch of people in a room, and gave them 3 light sources at approximately pure wavelengths (single wavelength for red, green, and blue). They asked them, given some "color" (light of a single wavelength along the visible spectrum), to adjust their 3 knobs until they perceived the color to match. As the wavelength to match is adjusted in many discrete increments, we get 3 RGB values for each participant. We can then take an average to get 3 fairly smooth curves.
Some (in this case, just red) knobs might need to be "negative". Since "amount of light" should be positive, we do this the same way we construct the integers - as pairs of positive numbers (their difference is the integer we're secretly representing). In other words, we add some, in this case red, light to the light we're trying to match until we can get a match with the remaining two.
The three color matching functions can be viewed as a line segment in 3 dimensional RGB space, parameterized by 380-700nm. To avoid negative values, we can choose a linear transformation (a 3x3 matrix) to map everything to the positive octant of "XYZ" space, which defines it. There's a fairly canonical one which brings RGB to unit vectors.
One of the directions will correspond to luminosity, or how bright the color is. The other two directions encode the chromaticity, or what we think of as a "color". To get the chromaticity, we essentially project onto the plane $R+G+B=1$. When we take convex combinations of points on the spectral line (the 380-700nm line segment of wavelengths we know we can represent with our RGBs), we fill in a 2d shape called the gamut, which looks a bit like a painter's palette.
Your question is essentially asking if white light (the centroid of the gamut) is along a line connecting two points, where one is in the subinterval "cyan" (490-520nm, though looks more like 490-500nm) and the other is in the subinterval "red" (630-700nm).
It looks like you can indeed add cyan and red to get white, but you need a bit more cyan than red.
One should be careful with notation for mixing colors. There are additive and subtractive ways of thinking about mixing colors.
Further, the intuitive use of the "+" symbol is a bit misleading. On the chromaticity diagram (or a 2d projection), when we mix colors we are generally taking an average (possibly weighted) . This operation is indeed commutative, and it is idempotent (red $\oplus$ red = red). However, it is not associative (I notice OP did put parentheses).
To get associativity, you have to include the luminosity, and use standard addition in 3d RGB space, which is commutative and associative but not idempotent.
If you want idempotency, commutativity, and associativity then the most general algebra (a free semi-lattice) on 3 discrete generators you can write down $\langle R,G,B \rangle$ turns out to have exactly 7 elements. The price you pay is invertibility. It is a Boolean algebra without the identity, which can be put in one-to-one correspondence with the Fano plane if you want. However, this algebra is not so mysterious. It's just the union (or intersection) algebra:
You can try mixing colors to play around (paint is also fun). The closest I could get to grey with this tool by mixing cyan and red was more a Payne's grey, a greyish blue: #7b7b84