Metamath Proof Explorer


Theorem cbvoprab2davw

Description: Change the second bound variable in an operation abstraction. Deduction form. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvoprab2davw.1 ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion cbvoprab2davw ( 𝜑 → { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∣ 𝜒 } )

Proof

Step Hyp Ref Expression
1 cbvoprab2davw.1 ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ( 𝜓 ↔ 𝜒 ) )
2 opeq2 ⊢ ( 𝑦 = 𝑤 → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝑥 , 𝑤 ⟩ )
3 2 adantl ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝑥 , 𝑤 ⟩ )
4 3 opeq1d ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ )
5 4 eqeq2d ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ↔ 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ) )
6 5 1 anbi12d ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ( ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
7 6 exbidv ⊢ ( ( 𝜑 ∧ 𝑦 = 𝑤 ) → ( ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
8 7 cbvexdvaw ⊢ ( 𝜑 → ( ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑤 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
9 8 exbidv ⊢ ( 𝜑 → ( ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑥 ∃ 𝑤 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
10 9 abbidv ⊢ ( 𝜑 → { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) } = { 𝑡 ∣ ∃ 𝑥 ∃ 𝑤 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) } )
11 df-oprab ⊢ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) }
12 df-oprab ⊢ { ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∣ 𝜒 } = { 𝑡 ∣ ∃ 𝑥 ∃ 𝑤 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) }
13 10 11 12 3eqtr4g ⊢ ( 𝜑 → { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { ⟨ ⟨ 𝑥 , 𝑤 ⟩ , 𝑧 ⟩ ∣ 𝜒 } )