Metamath Proof Explorer


Theorem cbvoprab1vw

Description: Change the first bound variable in an operation abstraction, using implicit substitution. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvoprab1vw.1 ⊢ ( 𝑥 = 𝑤 → ( 𝜓 ↔ 𝜒 ) )
Assertion cbvoprab1vw { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜒 }

Proof

Step Hyp Ref Expression
1 cbvoprab1vw.1 ⊢ ( 𝑥 = 𝑤 → ( 𝜓 ↔ 𝜒 ) )
2 opeq1 ⊢ ( 𝑥 = 𝑤 → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝑤 , 𝑦 ⟩ )
3 2 opeq1d ⊢ ( 𝑥 = 𝑤 → ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ )
4 3 eqeq2d ⊢ ( 𝑥 = 𝑤 → ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ↔ 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ) )
5 4 1 anbi12d ⊢ ( 𝑥 = 𝑤 → ( ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ( 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
6 5 2exbidv ⊢ ( 𝑥 = 𝑤 → ( ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) ) )
7 6 cbvexvw ⊢ ( ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) ↔ ∃ 𝑤 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) )
8 7 abbii ⊢ { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) } = { 𝑡 ∣ ∃ 𝑤 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) }
9 df-oprab ⊢ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { 𝑡 ∣ ∃ 𝑥 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜓 ) }
10 df-oprab ⊢ { ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜒 } = { 𝑡 ∣ ∃ 𝑤 ∃ 𝑦 ∃ 𝑧 ( 𝑡 = ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∧ 𝜒 ) }
11 8 9 10 3eqtr4i ⊢ { ⟨ ⟨ 𝑥 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜓 } = { ⟨ ⟨ 𝑤 , 𝑦 ⟩ , 𝑧 ⟩ ∣ 𝜒 }