Metamath Proof Explorer


Theorem ercpbl

Description: Translate the function compatibility relation to a quotient set. (Contributed by Mario Carneiro, 24-Feb-2015) (Revised by Mario Carneiro, 12-Aug-2015) (Revised by AV, 12-Jul-2024)

Ref Expression
Hypotheses ercpbl.r φ ˙ Er V
ercpbl.v φ V W
ercpbl.f F = x V x ˙
ercpbl.c φ a V b V a + ˙ b V
ercpbl.e φ A ˙ C B ˙ D A + ˙ B ˙ C + ˙ D
Assertion ercpbl φ A V B V C V D V F A = F C F B = F D F A + ˙ B = F C + ˙ D

Proof

Step Hyp Ref Expression
1 ercpbl.r φ ˙ Er V
2 ercpbl.v φ V W
3 ercpbl.f F = x V x ˙
4 ercpbl.c φ a V b V a + ˙ b V
5 ercpbl.e φ A ˙ C B ˙ D A + ˙ B ˙ C + ˙ D
6 5 3ad2ant1 φ A V B V C V D V A ˙ C B ˙ D A + ˙ B ˙ C + ˙ D
7 1 3ad2ant1 φ A V B V C V D V ˙ Er V
8 2 3ad2ant1 φ A V B V C V D V V W
9 simp2l φ A V B V C V D V A V
10 7 8 3 9 ercpbllem φ A V B V C V D V F A = F C A ˙ C
11 simp2r φ A V B V C V D V B V
12 7 8 3 11 ercpbllem φ A V B V C V D V F B = F D B ˙ D
13 10 12 anbi12d φ A V B V C V D V F A = F C F B = F D A ˙ C B ˙ D
14 4 caovclg φ A V B V A + ˙ B V
15 14 3adant3 φ A V B V C V D V A + ˙ B V
16 7 8 3 15 ercpbllem φ A V B V C V D V F A + ˙ B = F C + ˙ D A + ˙ B ˙ C + ˙ D
17 6 13 16 3imtr4d φ A V B V C V D V F A = F C F B = F D F A + ˙ B = F C + ˙ D