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Metamath Proof Explorer

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 ) )