Theorem sbievg

Description: Substitution applied to expressions linked by implicit substitution. The proof was part of a former cbvabw version. (Contributed by GG and WL, 26-Oct-2024)

	Ref	Expression
Hypotheses	sbievg.1	⊢ Ⅎ 𝑦 𝜑
	sbievg.2	⊢ Ⅎ 𝑥 𝜓
	sbievg.3	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
Assertion	sbievg	⊢ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 )

Proof

Step	Hyp	Ref	Expression
1		sbievg.1	⊢ Ⅎ 𝑦 𝜑
2		sbievg.2	⊢ Ⅎ 𝑥 𝜓
3		sbievg.3	⊢ ( 𝑥 = 𝑦 → ( 𝜑 ↔ 𝜓 ) )
4		nfv	⊢ Ⅎ 𝑦 𝑥 = 𝑤
5	4 1	nfim	⊢ Ⅎ 𝑦 ( 𝑥 = 𝑤 → 𝜑 )
6		nfv	⊢ Ⅎ 𝑥 𝑦 = 𝑤
7	6 2	nfim	⊢ Ⅎ 𝑥 ( 𝑦 = 𝑤 → 𝜓 )
8		equequ1	⊢ ( 𝑥 = 𝑦 → ( 𝑥 = 𝑤 ↔ 𝑦 = 𝑤 ) )
9	8 3	imbi12d	⊢ ( 𝑥 = 𝑦 → ( ( 𝑥 = 𝑤 → 𝜑 ) ↔ ( 𝑦 = 𝑤 → 𝜓 ) ) )
10	5 7 9	cbvalv1	⊢ ( ∀ 𝑥 ( 𝑥 = 𝑤 → 𝜑 ) ↔ ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜓 ) )
11	10	imbi2i	⊢ ( ( 𝑤 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑤 → 𝜑 ) ) ↔ ( 𝑤 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜓 ) ) )
12	11	albii	⊢ ( ∀ 𝑤 ( 𝑤 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑤 → 𝜑 ) ) ↔ ∀ 𝑤 ( 𝑤 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜓 ) ) )
13		df-sb	⊢ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ ∀ 𝑤 ( 𝑤 = 𝑧 → ∀ 𝑥 ( 𝑥 = 𝑤 → 𝜑 ) ) )
14		df-sb	⊢ ( [ 𝑧 / 𝑦 ] 𝜓 ↔ ∀ 𝑤 ( 𝑤 = 𝑧 → ∀ 𝑦 ( 𝑦 = 𝑤 → 𝜓 ) ) )
15	12 13 14	3bitr4i	⊢ ( [ 𝑧 / 𝑥 ] 𝜑 ↔ [ 𝑧 / 𝑦 ] 𝜓 )