Metamath Proof Explorer


Theorem cbvsbdavw

Description: Change bound variable in proper substitution. Deduction form. (Contributed by GG, 14-Aug-2025)

Ref Expression
Hypothesis cbvsbdavw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion cbvsbdavw ( 𝜑 → ( [ 𝑧 / 𝑥 ] 𝜓 ↔ [ 𝑧 / 𝑦 ] 𝜒 ) )

Proof

Step Hyp Ref Expression
1 cbvsbdavw.1 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝜓 ↔ 𝜒 ) )
2 equequ1 ⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑡 ↔ 𝑦 = 𝑡 ) )
3 2 adantl ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( 𝑥 = 𝑡 ↔ 𝑦 = 𝑡 ) )
4 3 1 imbi12d ⊢ ( ( 𝜑 ∧ 𝑥 = 𝑦 ) → ( ( 𝑥 = 𝑡 → 𝜓 ) ↔ ( 𝑦 = 𝑡 → 𝜒 ) ) )
5 4 cbvaldvaw ⊢ ( 𝜑 → ( ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) )
6 5 imbi2d ⊢ ( 𝜑 → ( ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) ) )
7 6 albidv ⊢ ( 𝜑 → ( ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) ) )
8 dfsb ⊢ ( [ 𝑧 / 𝑥 ] 𝜓 ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑡 → 𝜓 ) ) )
9 dfsb ⊢ ( [ 𝑧 / 𝑦 ] 𝜒 ↔ ∀ 𝑡 ( 𝑡 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑡 → 𝜒 ) ) )
10 7 8 9 3bitr4g ⊢ ( 𝜑 → ( [ 𝑧 / 𝑥 ] 𝜓 ↔ [ 𝑧 / 𝑦 ] 𝜒 ) )