Theorem sbceqg

Description: Distribute proper substitution through an equality relation. (Contributed by NM, 10-Nov-2005) (Proof shortened by Andrew Salmon, 29-Jun-2011)

	Ref	Expression
Assertion	sbceqg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		dfsbcq2	⊢ ( 𝑧 = 𝐴 → ( [ 𝑧 / 𝑥 ] 𝐵 = 𝐶 ↔ [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ) )
2		dfsbcq2	⊢ ( 𝑧 = 𝐴 → ( [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 ↔ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 ) )
3	2	abbidv	⊢ ( 𝑧 = 𝐴 → { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 } )
4		dfsbcq2	⊢ ( 𝑧 = 𝐴 → ( [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 ↔ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 ) )
5	4	abbidv	⊢ ( 𝑧 = 𝐴 → { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } )
6	3 5	eqeq12d	⊢ ( 𝑧 = 𝐴 → ( { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 } ↔ { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } ) )
7		nfs1v	⊢ Ⅎ 𝑥 [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵
8	7	nfab	⊢ Ⅎ 𝑥 { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 }
9		nfs1v	⊢ Ⅎ 𝑥 [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶
10	9	nfab	⊢ Ⅎ 𝑥 { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 }
11	8 10	nfeq	⊢ Ⅎ 𝑥 { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 }
12		sbab	⊢ ( 𝑥 = 𝑧 → 𝐵 = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } )
13		sbab	⊢ ( 𝑥 = 𝑧 → 𝐶 = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 } )
14	12 13	eqeq12d	⊢ ( 𝑥 = 𝑧 → ( 𝐵 = 𝐶 ↔ { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 } ) )
15	11 14	sbiev	⊢ ( [ 𝑧 / 𝑥 ] 𝐵 = 𝐶 ↔ { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝑧 / 𝑥 ] 𝑦 ∈ 𝐶 } )
16	1 6 15	vtoclbg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } ) )
17		df-csb	⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 }
18		df-csb	⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 }
19	17 18	eqeq12i	⊢ ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ↔ { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐵 } = { 𝑦 ∣ [ 𝐴 / 𝑥 ] 𝑦 ∈ 𝐶 } )
20	16 19	bitr4di	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ) )