Theorem sbcrel

Description: Distribute proper substitution through a relation predicate. (Contributed by Alexander van der Vekens, 23-Jul-2017)

	Ref	Expression
Assertion	sbcrel	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] Rel 𝑅 ↔ Rel ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcssg	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝑅 ⊆ ( V × V ) ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ⊆ ⦋ 𝐴 / 𝑥 ⦌ ( V × V ) ) )
2		csbconstg	⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( V × V ) = ( V × V ) )
3	2	sseq2d	⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ⊆ ⦋ 𝐴 / 𝑥 ⦌ ( V × V ) ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ⊆ ( V × V ) ) )
4	1 3	bitrd	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝑅 ⊆ ( V × V ) ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ⊆ ( V × V ) ) )
5		df-rel	⊢ ( Rel 𝑅 ↔ 𝑅 ⊆ ( V × V ) )
6	5	sbcbii	⊢ ( [ 𝐴 / 𝑥 ] Rel 𝑅 ↔ [ 𝐴 / 𝑥 ] 𝑅 ⊆ ( V × V ) )
7		df-rel	⊢ ( Rel ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ⊆ ( V × V ) )
8	4 6 7	3bitr4g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] Rel 𝑅 ↔ Rel ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ) )