Metamath Proof Explorer


Theorem csbiebg

Description: Bidirectional conversion between an implicit class substitution hypothesis x = A -> B = C and its explicit substitution equivalent. (Contributed by NM, 24-Mar-2013) (Revised by Mario Carneiro, 11-Dec-2016)

Ref Expression
Hypothesis csbiebg.2 ⊢ Ⅎ 𝑥 𝐶
Assertion csbiebg ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝐵 = 𝐶 ) ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 csbiebg.2 ⊢ Ⅎ 𝑥 𝐶
2 eqeq2 ⊢ ( 𝑎 = 𝐴 → ( 𝑥 = 𝑎 ↔ 𝑥 = 𝐴 ) )
3 2 imbi1d ⊢ ( 𝑎 = 𝐴 → ( ( 𝑥 = 𝑎 → 𝐵 = 𝐶 ) ↔ ( 𝑥 = 𝐴 → 𝐵 = 𝐶 ) ) )
4 3 albidv ⊢ ( 𝑎 = 𝐴 → ( ∀ 𝑥 ( 𝑥 = 𝑎 → 𝐵 = 𝐶 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴 → 𝐵 = 𝐶 ) ) )
5 csbeq1 ⊢ ( 𝑎 = 𝐴 → ⦋ 𝑎 / 𝑥 ⦌ 𝐵 = ⦋ 𝐴 / 𝑥 ⦌ 𝐵 )
6 5 eqeq1d ⊢ ( 𝑎 = 𝐴 → ( ⦋ 𝑎 / 𝑥 ⦌ 𝐵 = 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )
7 vex ⊢ 𝑎 ∈ V
8 7 1 csbieb ⊢ ( ∀ 𝑥 ( 𝑥 = 𝑎 → 𝐵 = 𝐶 ) ↔ ⦋ 𝑎 / 𝑥 ⦌ 𝐵 = 𝐶 )
9 4 6 8 vtoclbg ⊢ ( 𝐴 ∈ 𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴 → 𝐵 = 𝐶 ) ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 = 𝐶 ) )