# Metamath Proof Explorer

## Theorem eqeq1d

Description: Deduction from equality to equivalence of equalities. (Contributed by NM, 27-Dec-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 5-Dec-2019)

Ref Expression
Hypothesis eqeq1d.1
`|- ( ph -> A = B )`
Assertion eqeq1d
`|- ( ph -> ( A = C <-> B = C ) )`

### Proof

Step Hyp Ref Expression
1 eqeq1d.1
` |-  ( ph -> A = B )`
2 dfcleq
` |-  ( A = B <-> A. x ( x e. A <-> x e. B ) )`
3 2 biimpi
` |-  ( A = B -> A. x ( x e. A <-> x e. B ) )`
4 bibi1
` |-  ( ( x e. A <-> x e. B ) -> ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) )`
5 4 alimi
` |-  ( A. x ( x e. A <-> x e. B ) -> A. x ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) )`
6 albi
` |-  ( A. x ( ( x e. A <-> x e. C ) <-> ( x e. B <-> x e. C ) ) -> ( A. x ( x e. A <-> x e. C ) <-> A. x ( x e. B <-> x e. C ) ) )`
7 1 3 5 6 4syl
` |-  ( ph -> ( A. x ( x e. A <-> x e. C ) <-> A. x ( x e. B <-> x e. C ) ) )`
8 dfcleq
` |-  ( A = C <-> A. x ( x e. A <-> x e. C ) )`
9 dfcleq
` |-  ( B = C <-> A. x ( x e. B <-> x e. C ) )`
10 7 8 9 3bitr4g
` |-  ( ph -> ( A = C <-> B = C ) )`