Metamath Proof Explorer


Theorem cphreccllem

Description: Lemma for cphreccl . (Contributed by Mario Carneiro, 8-Oct-2015)

Ref Expression
Hypotheses cphsubrglem.k K = Base F
cphsubrglem.1 φ F = fld 𝑠 A
cphsubrglem.2 φ F DivRing
Assertion cphreccllem φ X K X 0 1 X K

Proof

Step Hyp Ref Expression
1 cphsubrglem.k K = Base F
2 cphsubrglem.1 φ F = fld 𝑠 A
3 cphsubrglem.2 φ F DivRing
4 1 2 3 cphsubrglem φ F = fld 𝑠 K K = A K SubRing fld
5 4 simp3d φ K SubRing fld
6 5 3ad2ant1 φ X K X 0 K SubRing fld
7 cnfldbas = Base fld
8 7 subrgss K SubRing fld K
9 6 8 syl φ X K X 0 K
10 simp2 φ X K X 0 X K
11 9 10 sseldd φ X K X 0 X
12 simp3 φ X K X 0 X 0
13 cnfldinv X X 0 inv r fld X = 1 X
14 11 12 13 syl2anc φ X K X 0 inv r fld X = 1 X
15 eqid fld 𝑠 K = fld 𝑠 K
16 cnfld0 0 = 0 fld
17 15 16 subrg0 K SubRing fld 0 = 0 fld 𝑠 K
18 6 17 syl φ X K X 0 0 = 0 fld 𝑠 K
19 4 simp1d φ F = fld 𝑠 K
20 19 3ad2ant1 φ X K X 0 F = fld 𝑠 K
21 20 fveq2d φ X K X 0 0 F = 0 fld 𝑠 K
22 18 21 eqtr4d φ X K X 0 0 = 0 F
23 12 22 neeqtrd φ X K X 0 X 0 F
24 eldifsn X K 0 F X K X 0 F
25 10 23 24 sylanbrc φ X K X 0 X K 0 F
26 3 3ad2ant1 φ X K X 0 F DivRing
27 eqid Unit F = Unit F
28 eqid 0 F = 0 F
29 1 27 28 isdrng F DivRing F Ring Unit F = K 0 F
30 29 simprbi F DivRing Unit F = K 0 F
31 26 30 syl φ X K X 0 Unit F = K 0 F
32 20 fveq2d φ X K X 0 Unit F = Unit fld 𝑠 K
33 31 32 eqtr3d φ X K X 0 K 0 F = Unit fld 𝑠 K
34 25 33 eleqtrd φ X K X 0 X Unit fld 𝑠 K
35 eqid Unit fld = Unit fld
36 eqid Unit fld 𝑠 K = Unit fld 𝑠 K
37 eqid inv r fld = inv r fld
38 15 35 36 37 subrgunit K SubRing fld X Unit fld 𝑠 K X Unit fld X K inv r fld X K
39 6 38 syl φ X K X 0 X Unit fld 𝑠 K X Unit fld X K inv r fld X K
40 34 39 mpbid φ X K X 0 X Unit fld X K inv r fld X K
41 40 simp3d φ X K X 0 inv r fld X K
42 14 41 eqeltrrd φ X K X 0 1 X K