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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 ) )