Metamath Proof Explorer


Theorem bnj1501

Description: Technical lemma for bnj1500 . This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj1501.1 B = d | d A x d pred x A R d
bnj1501.2 Y = x f pred x A R
bnj1501.3 C = f | d B f Fn d x d f x = G Y
bnj1501.4 F = C
bnj1501.5 φ R FrSe A x A
bnj1501.6 ψ φ f C x dom f
bnj1501.7 χ ψ d B dom f = d
Assertion bnj1501 R FrSe A x A F x = G x F pred x A R

Proof

Step Hyp Ref Expression
1 bnj1501.1 B = d | d A x d pred x A R d
2 bnj1501.2 Y = x f pred x A R
3 bnj1501.3 C = f | d B f Fn d x d f x = G Y
4 bnj1501.4 F = C
5 bnj1501.5 φ R FrSe A x A
6 bnj1501.6 ψ φ f C x dom f
7 bnj1501.7 χ ψ d B dom f = d
8 5 simprbi φ x A
9 1 2 3 4 bnj60 R FrSe A F Fn A
10 9 fndmd R FrSe A dom F = A
11 5 10 bnj832 φ dom F = A
12 8 11 eleqtrrd φ x dom F
13 4 dmeqi dom F = dom C
14 3 bnj1317 w C f w C
15 14 bnj1400 dom C = f C dom f
16 13 15 eqtri dom F = f C dom f
17 12 16 eleqtrdi φ x f C dom f
18 17 bnj1405 φ f C x dom f
19 18 6 bnj1209 φ f ψ
20 3 bnj1436 f C d B f Fn d x d f x = G Y
21 20 bnj1299 f C d B f Fn d
22 fndm f Fn d dom f = d
23 21 22 bnj31 f C d B dom f = d
24 6 23 bnj836 ψ d B dom f = d
25 1 2 3 4 5 6 bnj1518 ψ d ψ
26 24 7 25 bnj1521 ψ d χ
27 9 fnfund R FrSe A Fun F
28 5 27 bnj832 φ Fun F
29 6 28 bnj835 ψ Fun F
30 elssuni f C f C
31 30 4 sseqtrrdi f C f F
32 6 31 bnj836 ψ f F
33 6 simp3bi ψ x dom f
34 29 32 33 bnj1502 ψ F x = f x
35 1 2 3 bnj1514 f C x dom f f x = G Y
36 6 35 bnj836 ψ x dom f f x = G Y
37 36 33 bnj1294 ψ f x = G Y
38 34 37 eqtrd ψ F x = G Y
39 7 38 bnj835 χ F x = G Y
40 7 29 bnj835 χ Fun F
41 7 32 bnj835 χ f F
42 1 bnj1517 d B x d pred x A R d
43 7 42 bnj836 χ x d pred x A R d
44 7 33 bnj835 χ x dom f
45 7 simp3bi χ dom f = d
46 44 45 eleqtrd χ x d
47 43 46 bnj1294 χ pred x A R d
48 47 45 sseqtrrd χ pred x A R dom f
49 40 41 48 bnj1503 χ F pred x A R = f pred x A R
50 49 opeq2d χ x F pred x A R = x f pred x A R
51 50 2 eqtr4di χ x F pred x A R = Y
52 51 fveq2d χ G x F pred x A R = G Y
53 39 52 eqtr4d χ F x = G x F pred x A R
54 26 53 bnj593 ψ d F x = G x F pred x A R
55 1 2 3 4 bnj1519 F x = G x F pred x A R d F x = G x F pred x A R
56 54 55 bnj1397 ψ F x = G x F pred x A R
57 19 56 bnj593 φ f F x = G x F pred x A R
58 1 2 3 4 bnj1520 F x = G x F pred x A R f F x = G x F pred x A R
59 57 58 bnj1397 φ F x = G x F pred x A R
60 5 59 bnj1459 R FrSe A x A F x = G x F pred x A R