Theorem sbrimvlem

Description: Common proof template for sbrimvw and sbrimv . The hypothesis is an instance of 19.21 . (Contributed by Wolf Lammen, 29-Jan-2024)

	Ref	Expression
Hypothesis	sbrimvlem.1	⊢ ( ∀ 𝑥 ( 𝜑 → ( 𝑥 = 𝑦 → 𝜓 ) ) ↔ ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ) )
Assertion	sbrimvlem	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )

Proof

Step	Hyp	Ref	Expression
1		sbrimvlem.1	⊢ ( ∀ 𝑥 ( 𝜑 → ( 𝑥 = 𝑦 → 𝜓 ) ) ↔ ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ) )
2		sb6	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑 → 𝜓 ) ) )
3		bi2.04	⊢ ( ( 𝜑 → ( 𝑥 = 𝑦 → 𝜓 ) ) ↔ ( 𝑥 = 𝑦 → ( 𝜑 → 𝜓 ) ) )
4	3	albii	⊢ ( ∀ 𝑥 ( 𝜑 → ( 𝑥 = 𝑦 → 𝜓 ) ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → ( 𝜑 → 𝜓 ) ) )
5	2 4 1	3bitr2i	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ) )
6		sb6	⊢ ( [ 𝑦 / 𝑥 ] 𝜓 ↔ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) )
7	6	imbi2i	⊢ ( ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) ↔ ( 𝜑 → ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜓 ) ) )
8	5 7	bitr4i	⊢ ( [ 𝑦 / 𝑥 ] ( 𝜑 → 𝜓 ) ↔ ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜓 ) )