= Solution
Whenever $\alpha$ and $\beta$ bound a bigon, isotope one side across the bigon. The two removed crossings have opposite local signs: the induced directions around the two corners of an oriented disc are opposite. Thus the move reduces the <geometric intersection number> by two while leaving the <algebraic intersection number of curves on an oriented surface> unchanged.
Repeatedly remove bigons. The process terminates because the intersection count is a nonnegative integer, and the <bigon criterion> says that the resulting curves $\alpha_0$ and $\beta_0$ are in minimal position. Since every removal cancelled one positive and one negative crossing,
$$
\langle\alpha,\beta\rangle=\langle\alpha_0,\beta_0\rangle.
$$
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