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Metamath Proof Explorer

Theorem mpoeq3dva

Description: Slightly more general equality inference for the maps-to notation. (Contributed by NM, 17-Oct-2013)

Ref Expression
Hypothesis mpoeq3dva.1 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → 𝐶 = 𝐷 )
Assertion mpoeq3dva ( 𝜑 → ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐶 ) = ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 mpoeq3dva.1 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) → 𝐶 = 𝐷 )
2 1 3expb ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ) → 𝐶 = 𝐷 )
3 2 eqeq2d ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑧 = 𝐶 ↔ 𝑧 = 𝐷 ) )
4 3 pm5.32da ⊢ ( 𝜑 → ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐶 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐷 ) ) )
5 4 oprabbidv ⊢ ( 𝜑 → { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐶 ) } = { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐷 ) } )
6 df-mpo ⊢ ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐶 ) = { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐶 ) }
7 df-mpo ⊢ ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐷 ) = { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝑧 = 𝐷 ) }
8 5 6 7 3eqtr4g ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐶 ) = ( 𝑥 ∈ 𝐴 , 𝑦 ∈ 𝐵 ↦ 𝐷 ) )