Metamath Proof Explorer


Theorem cbvmptdavw

Description: Change bound variable in a maps-to function. Deduction form. (Contributed by GG, 14-Aug-2025)

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

Proof

Step Hyp Ref Expression
1 cbvmptdavw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → 𝐵 = 𝐶 )
2 eleq1w ⊢ ( 𝑥 = 𝑦 → ( 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 ) )
3 2 adantl ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴 ) )
4 1 eqeq2d ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝑡 = 𝐵 ↔ 𝑡 = 𝐶 ) )
5 3 4 anbi12d ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( ( 𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵 ) ↔ ( 𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶 ) ) )
6 5 cbvopab1davw ⊢ ( 𝜑 → { ⟨ 𝑥 , 𝑡 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵 ) } = { ⟨ 𝑦 , 𝑡 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶 ) } )
7 df-mpt ⊢ ( 𝑥 ∈ 𝐴 ↦ 𝐵 ) = { ⟨ 𝑥 , 𝑡 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑡 = 𝐵 ) }
8 df-mpt ⊢ ( 𝑦 ∈ 𝐴 ↦ 𝐶 ) = { ⟨ 𝑦 , 𝑡 ⟩ ∣ ( 𝑦 ∈ 𝐴 ∧ 𝑡 = 𝐶 ) }
9 6 7 8 3eqtr4g ⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↦ 𝐵 ) = ( 𝑦 ∈ 𝐴 ↦ 𝐶 ) )