Metamath Proof Explorer


Theorem elxpi

Description: Membership in a Cartesian product. Uses fewer axioms than elxp . (Contributed by NM, 4-Jul-1994)

Ref Expression
Assertion elxpi ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → ∃ 𝑥𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 eqeq1 ( 𝑧 = 𝐴 → ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ↔ 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ) )
2 1 anbi1d ( 𝑧 = 𝐴 → ( ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) ↔ ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) ) )
3 2 2exbidv ( 𝑧 = 𝐴 → ( ∃ 𝑥𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) ↔ ∃ 𝑥𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) ) )
4 df-xp ( 𝐵 × 𝐶 ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐵𝑦𝐶 ) }
5 df-opab { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐵𝑦𝐶 ) } = { 𝑧 ∣ ∃ 𝑥𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) }
6 4 5 eqtri ( 𝐵 × 𝐶 ) = { 𝑧 ∣ ∃ 𝑥𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) }
7 3 6 elab2g ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → ( 𝐴 ∈ ( 𝐵 × 𝐶 ) ↔ ∃ 𝑥𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) ) )
8 7 ibi ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → ∃ 𝑥𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥𝐵𝑦𝐶 ) ) )