Derivation of Snell's Law

I am following an example in the book "Electromagnetic Waves and Antennas" by S. Orfanidis. He has two planes waves incident on both sides of a boundary. Here, $k_{pm} = hat{x} k_{x pm} + hat{y} k_{y pm} + hat{z} k_{z pm} $ is the forward and backward propagation vector for the left medium and $k’_{pm} = hat{x} k’_{x pm} + hat{y} k’_{y pm} + hat{z} k’_{z pm} $ is the forward and backward propagation vector for the right medium. For the boundary condition at z=0 he writes

$vec{E}_+ e^{-j(k_{x+}x + k_{y+}y)} + vec{E}_- e^{-j(k_{x-}x + k_{y-}y)} = vec{E’}_+ e^{-j(k’_{x+}x + k’_{y+}y)} + vec{E’}_- e^{-j(k’_{x-}x + k’_{y-}y)}$

where the $vec{E}$ ‘s are the tangential component of the electric fields and the prime denotes the second medium.

I understand that for all points on the boundary, LHS=RHS needs to hold. What I don’t find so obvious is that from this he says,

$ e^{-j(k_{x+}x + k_{y+}y)} = e^{-j(k_{x-}x + k_{y-}y)} = e^{-j(k’_{x+}x + k’_{y+}y)} = e^{-j(k’_{x-}x + k’_{y-}y)} $.

How would we prove this?

Subsequently, I don’t see how $k_{x+} = k_{x-} = k’_{x+} = k’_{x-}$ and $k_{y+} = k_{y-} = k’_{y+} = k’_{y-}$ can be derived from this.