Metamath Proof Explorer


Theorem finxpreclem5

Description: Lemma for ^^ recursion theorems. (Contributed by ML, 24-Oct-2020)

Ref Expression
Hypothesis finxpreclem5.1 F=nω,xVifn=1𝑜xUifxV×Un1stxnx
Assertion finxpreclem5 nω1𝑜n¬xV×UFnx=nx

Proof

Step Hyp Ref Expression
1 finxpreclem5.1 F=nω,xVifn=1𝑜xUifxV×Un1stxnx
2 df-ov nFx=Fnx
3 vex xV
4 0ex V
5 opex n1stxV
6 opex nxV
7 5 6 ifex ifxV×Un1stxnxV
8 4 7 ifex ifn=1𝑜xUifxV×Un1stxnxV
9 1 ovmpt4g nωxVifn=1𝑜xUifxV×Un1stxnxVnFx=ifn=1𝑜xUifxV×Un1stxnx
10 3 8 9 mp3an23 nωnFx=ifn=1𝑜xUifxV×Un1stxnx
11 10 ad2antrr nω1𝑜n¬xV×UnFx=ifn=1𝑜xUifxV×Un1stxnx
12 1on 1𝑜On
13 12 onirri ¬1𝑜1𝑜
14 eleq2 n=1𝑜1𝑜n1𝑜1𝑜
15 13 14 mtbiri n=1𝑜¬1𝑜n
16 15 con2i 1𝑜n¬n=1𝑜
17 16 intnanrd 1𝑜n¬n=1𝑜xU
18 17 iffalsed 1𝑜nifn=1𝑜xUifxV×Un1stxnx=ifxV×Un1stxnx
19 18 adantl nω1𝑜nifn=1𝑜xUifxV×Un1stxnx=ifxV×Un1stxnx
20 iffalse ¬xV×UifxV×Un1stxnx=nx
21 19 20 sylan9eq nω1𝑜n¬xV×Uifn=1𝑜xUifxV×Un1stxnx=nx
22 11 21 eqtrd nω1𝑜n¬xV×UnFx=nx
23 2 22 eqtr3id nω1𝑜n¬xV×UFnx=nx
24 23 ex nω1𝑜n¬xV×UFnx=nx