Metamath Proof Explorer


Theorem cbvopab1davw

Description: Change the first bound variable in an ordered-pair class abstraction. Deduction form. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvopab1davw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion cbvopab1davw ( 𝜑 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } = { ⟨ 𝑧 , 𝑦 ⟩ ∣ 𝜒 } )

Proof

Step Hyp Ref Expression
1 cbvopab1davw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ( 𝜓 ↔ 𝜒 ) )
2 opeq1 ⊢ ( 𝑥 = 𝑧 → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝑧 , 𝑦 ⟩ )
3 2 adantl ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝑧 , 𝑦 ⟩ )
4 3 eqeq2d ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ↔ 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ) )
5 4 1 anbi12d ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ( ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ↔ ( 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ∧ 𝜒 ) ) )
6 5 exbidv ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑧 ) → ( ∃ 𝑦 ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑦 ( 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ∧ 𝜒 ) ) )
7 6 cbvexdvaw ⊢ ( 𝜑 → ( ∃ 𝑥 ∃ 𝑦 ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑧 ∃ 𝑦 ( 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ∧ 𝜒 ) ) )
8 7 abbidv ⊢ ( 𝜑 → { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) } = { 𝑡 ∣ ∃ 𝑧 ∃ 𝑦 ( 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ∧ 𝜒 ) } )
9 df-opab ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } = { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑡 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜓 ) }
10 df-opab ⊢ { ⟨ 𝑧 , 𝑦 ⟩ ∣ 𝜒 } = { 𝑡 ∣ ∃ 𝑧 ∃ 𝑦 ( 𝑡 = ⟨ 𝑧 , 𝑦 ⟩ ∧ 𝜒 ) }
11 8 9 10 3eqtr4g ⊢ ( 𝜑 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜓 } = { ⟨ 𝑧 , 𝑦 ⟩ ∣ 𝜒 } )