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Proof.
Pick a common determination 𝒳 \mathcal{X} of f , g f,g and the θ i \theta_{i} ’s, and
divisors D , D ′ , D i D,D^{\prime},D_{i} such that f = φ D f=\varphi_{D} , g = φ D ′ g=\varphi_{D^{\prime}} and θ i \theta_{i} is the class
in N 1 ( 𝒳 / S ) N^{1}(\mathcal{X}/S) induced by D i D_{i} . Then by definition we have
∫ f d d c g ∧ θ 1 ∧ ⋯ ∧ θ n − 1 = ∑ E ord E ( D ) ( D ′ | E ⋅ D 1 | E ⋅ … ⋅ D n − 1 | E ) = ∑ E , E ′ ord E ( D ) ord E ′ ( D ′ ) ( D 1 | E ) | E ′ ∩ E ⋅ … ⋅ ( D n − 1 | E ) | E ′ ∩ E = ∑ E , E ′ ord E ( D ) ord E ′ ( D ′ ) ( D 1 | E ′ ) | E ∩ E ′ ⋅ … ⋅ ( D n − 1 | E ′ ) | E ∩ E ′ = ∫ g d d c f ∧ θ 1 ∧ ⋯ ∧ θ n − 1 \int f\,dd^{c}g\wedge\theta_{1}\wedge\dots\wedge\theta_{n-1}=\sum_{E}\ord_{E}(D)(D^{\prime}|_{E}\cdot D_{1}|_{E}\cdot...\cdot D_{n-1}|_{E})\\
=\sum_{E,E^{\prime}}\ord_{E}(D)\ord_{E^{\prime}}(D^{\prime})\,(D_{1}|_{E})|_{E^{\prime}\cap E}\cdot...\cdot(D_{n-1}|_{E})|_{E^{\prime}\cap E}\\
=\sum_{E,E^{\prime}}\ord_{E}(D)\ord_{E^{\prime}}(D^{\prime})\,(D_{1}|_{E^{\prime}})|_{E\cap E^{\prime}}\cdot...\cdot(D_{n-1}|_{E^{\prime}})|_{E\cap E^{\prime}}\\
=\int g\,dd^{c}f\wedge\theta_{1}\wedge\dots\wedge\theta_{n-1}
where the third equality follows from [Ful98 , Theorem 2.4] .
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