Metamath Proof Explorer


Theorem sspred

Description: Another subset/predecessor class relationship. (Contributed by Scott Fenton, 6-Feb-2011)

Ref Expression
Assertion sspred B A Pred R A X B Pred R A X = Pred R B X

Proof

Step Hyp Ref Expression
1 sseqin2 B A A B = B
2 df-pred Pred R A X = A R -1 X
3 2 sseq1i Pred R A X B A R -1 X B
4 df-ss A R -1 X B A R -1 X B = A R -1 X
5 in32 A R -1 X B = A B R -1 X
6 5 eqeq1i A R -1 X B = A R -1 X A B R -1 X = A R -1 X
7 3 4 6 3bitri Pred R A X B A B R -1 X = A R -1 X
8 ineq1 A B = B A B R -1 X = B R -1 X
9 8 eqeq1d A B = B A B R -1 X = A R -1 X B R -1 X = A R -1 X
10 9 biimpa A B = B A B R -1 X = A R -1 X B R -1 X = A R -1 X
11 df-pred Pred R B X = B R -1 X
12 10 11 2 3eqtr4g A B = B A B R -1 X = A R -1 X Pred R B X = Pred R A X
13 12 eqcomd A B = B A B R -1 X = A R -1 X Pred R A X = Pred R B X
14 1 7 13 syl2anb B A Pred R A X B Pred R A X = Pred R B X