Metamath Proof Explorer


Theorem acongeq12d

Description: Substitution deduction for alternating congruence. (Contributed by Stefan O'Rear, 3-Oct-2014)

Ref Expression
Hypotheses acongeq12d.1
|- ( ph -> B = C )
acongeq12d.2
|- ( ph -> D = E )
Assertion acongeq12d
|- ( ph -> ( ( A || ( B - D ) \/ A || ( B - -u D ) ) <-> ( A || ( C - E ) \/ A || ( C - -u E ) ) ) )

Proof

Step Hyp Ref Expression
1 acongeq12d.1
 |-  ( ph -> B = C )
2 acongeq12d.2
 |-  ( ph -> D = E )
3 1 2 oveq12d
 |-  ( ph -> ( B - D ) = ( C - E ) )
4 3 breq2d
 |-  ( ph -> ( A || ( B - D ) <-> A || ( C - E ) ) )
5 2 negeqd
 |-  ( ph -> -u D = -u E )
6 1 5 oveq12d
 |-  ( ph -> ( B - -u D ) = ( C - -u E ) )
7 6 breq2d
 |-  ( ph -> ( A || ( B - -u D ) <-> A || ( C - -u E ) ) )
8 4 7 orbi12d
 |-  ( ph -> ( ( A || ( B - D ) \/ A || ( B - -u D ) ) <-> ( A || ( C - E ) \/ A || ( C - -u E ) ) ) )