Metamath Proof Explorer


Theorem imasubc3lem2

Description: Lemma for imasubc3 . (Contributed by Zhi Wang, 7-Nov-2025)

Ref Expression
Hypotheses imasubc3lem1.s S = F A
imasubc3lem1.f φ F : B 1-1 C
imasubc3lem1.x φ X S
imasubc3lem2.y φ Y S
imasubc3lem2.f φ F V
imasubc3lem2.k K = x S , y S p F -1 x × F -1 y G p H p
Assertion imasubc3lem2 φ X K Y = F -1 X G F -1 Y F -1 X H F -1 Y

Proof

Step Hyp Ref Expression
1 imasubc3lem1.s S = F A
2 imasubc3lem1.f φ F : B 1-1 C
3 imasubc3lem1.x φ X S
4 imasubc3lem2.y φ Y S
5 imasubc3lem2.f φ F V
6 imasubc3lem2.k K = x S , y S p F -1 x × F -1 y G p H p
7 5 5 3 4 6 imasubclem3 φ X K Y = p F -1 X × F -1 Y G p H p
8 1 2 3 imasubc3lem1 φ F -1 X = F -1 X F F -1 X = X F -1 X B
9 8 simp1d φ F -1 X = F -1 X
10 1 2 4 imasubc3lem1 φ F -1 Y = F -1 Y F F -1 Y = Y F -1 Y B
11 10 simp1d φ F -1 Y = F -1 Y
12 9 11 xpeq12d φ F -1 X × F -1 Y = F -1 X × F -1 Y
13 fvex F -1 X V
14 fvex F -1 Y V
15 13 14 xpsn F -1 X × F -1 Y = F -1 X F -1 Y
16 12 15 eqtr3di φ F -1 X × F -1 Y = F -1 X F -1 Y
17 16 iuneq1d φ p F -1 X × F -1 Y G p H p = p F -1 X F -1 Y G p H p
18 7 17 eqtrd φ X K Y = p F -1 X F -1 Y G p H p
19 opex F -1 X F -1 Y V
20 fveq2 p = F -1 X F -1 Y G p = G F -1 X F -1 Y
21 df-ov F -1 X G F -1 Y = G F -1 X F -1 Y
22 20 21 eqtr4di p = F -1 X F -1 Y G p = F -1 X G F -1 Y
23 fveq2 p = F -1 X F -1 Y H p = H F -1 X F -1 Y
24 df-ov F -1 X H F -1 Y = H F -1 X F -1 Y
25 23 24 eqtr4di p = F -1 X F -1 Y H p = F -1 X H F -1 Y
26 22 25 imaeq12d p = F -1 X F -1 Y G p H p = F -1 X G F -1 Y F -1 X H F -1 Y
27 19 26 iunxsn p F -1 X F -1 Y G p H p = F -1 X G F -1 Y F -1 X H F -1 Y
28 18 27 eqtrdi φ X K Y = F -1 X G F -1 Y F -1 X H F -1 Y