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 dfss2 ⊢ 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