Metamath Proof Explorer


Theorem eqsbc1

Description: Substitution for the left-hand side in an equality. Class version of eqsb1 . (Contributed by Andrew Salmon, 29-Jun-2011) Avoid ax-13 . (Revised by Wolf Lammen, 29-Apr-2023)

Ref Expression
Assertion eqsbc1 ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐵 ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 dfsbcq ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐵 ↔ [ 𝐴 / 𝑥 ] 𝑥 = 𝐵 ) )
2 eqeq1 ⊢ ( 𝑦 = 𝐴 → ( 𝑦 = 𝐵 ↔ 𝐴 = 𝐵 ) )
3 sbsbc ⊢ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐵 ↔ [ 𝑦 / 𝑥 ] 𝑥 = 𝐵 )
4 eqsb1 ⊢ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐵 ↔ 𝑦 = 𝐵 )
5 3 4 bitr3i ⊢ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝐵 ↔ 𝑦 = 𝐵 )
6 1 2 5 vtoclbg ⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝑥 = 𝐵 ↔ 𝐴 = 𝐵 ) )