Metamath Proof Explorer


Theorem metakunt8

Description: C is the left inverse for A. (Contributed by metakunt, 24-May-2024)

Ref Expression
Hypotheses metakunt8.1 φM
metakunt8.2 φI
metakunt8.3 φIM
metakunt8.4 A=x1Mifx=IMifx<Ixx1
metakunt8.5 C=y1Mify=MIify<Iyy+1
metakunt8.6 φX1M
Assertion metakunt8 φI<XCAX=X

Proof

Step Hyp Ref Expression
1 metakunt8.1 φM
2 metakunt8.2 φI
3 metakunt8.3 φIM
4 metakunt8.4 A=x1Mifx=IMifx<Ixx1
5 metakunt8.5 C=y1Mify=MIify<Iyy+1
6 metakunt8.6 φX1M
7 5 a1i φI<XC=y1Mify=MIify<Iyy+1
8 eqeq1 y=AXy=MAX=M
9 breq1 y=AXy<IAX<I
10 id y=AXy=AX
11 oveq1 y=AXy+1=AX+1
12 9 10 11 ifbieq12d y=AXify<Iyy+1=ifAX<IAXAX+1
13 8 12 ifbieq2d y=AXify=MIify<Iyy+1=ifAX=MIifAX<IAXAX+1
14 13 adantl φI<Xy=AXify=MIify<Iyy+1=ifAX=MIifAX<IAXAX+1
15 1 2 3 4 5 6 metakunt7 φI<XAX=X1¬AX=M¬AX<I
16 15 simp2d φI<X¬AX=M
17 iffalse ¬AX=MifAX=MIifAX<IAXAX+1=ifAX<IAXAX+1
18 16 17 syl φI<XifAX=MIifAX<IAXAX+1=ifAX<IAXAX+1
19 15 simp3d φI<X¬AX<I
20 iffalse ¬AX<IifAX<IAXAX+1=AX+1
21 19 20 syl φI<XifAX<IAXAX+1=AX+1
22 18 21 eqtrd φI<XifAX=MIifAX<IAXAX+1=AX+1
23 15 simp1d φI<XAX=X1
24 23 oveq1d φI<XAX+1=X-1+1
25 elfznn X1MX
26 6 25 syl φX
27 26 nncnd φX
28 1cnd φ1
29 27 28 npcand φX-1+1=X
30 29 adantr φI<XX-1+1=X
31 24 30 eqtrd φI<XAX+1=X
32 22 31 eqtrd φI<XifAX=MIifAX<IAXAX+1=X
33 32 adantr φI<Xy=AXifAX=MIifAX<IAXAX+1=X
34 14 33 eqtrd φI<Xy=AXify=MIify<Iyy+1=X
35 1 2 3 4 metakunt1 φA:1M1M
36 35 adantr φI<XA:1M1M
37 6 adantr φI<XX1M
38 36 37 ffvelcdmd φI<XAX1M
39 7 34 38 37 fvmptd φI<XCAX=X