Metamath Proof Explorer


Theorem nnaordex

Description: Equivalence for ordering. Compare Exercise 23 of Enderton p. 88. (Contributed by NM, 5-Dec-1995) (Revised by Mario Carneiro, 15-Nov-2014)

Ref Expression
Assertion nnaordex AωBωABxωxA+𝑜x=B

Proof

Step Hyp Ref Expression
1 nnon BωBOn
2 1 adantl AωBωBOn
3 onelss BOnABAB
4 2 3 syl AωBωABAB
5 nnawordex AωBωABxωA+𝑜x=B
6 4 5 sylibd AωBωABxωA+𝑜x=B
7 simplr AωABxωAB
8 eleq2 A+𝑜x=BAA+𝑜xAB
9 7 8 syl5ibrcom AωABxωA+𝑜x=BAA+𝑜x
10 peano1 ω
11 nnaord ωxωAωxA+𝑜A+𝑜x
12 10 11 mp3an1 xωAωxA+𝑜A+𝑜x
13 12 ancoms AωxωxA+𝑜A+𝑜x
14 nna0 AωA+𝑜=A
15 14 adantr AωxωA+𝑜=A
16 15 eleq1d AωxωA+𝑜A+𝑜xAA+𝑜x
17 13 16 bitrd AωxωxAA+𝑜x
18 17 adantlr AωABxωxAA+𝑜x
19 9 18 sylibrd AωABxωA+𝑜x=Bx
20 19 ancrd AωABxωA+𝑜x=BxA+𝑜x=B
21 20 reximdva AωABxωA+𝑜x=BxωxA+𝑜x=B
22 21 ex AωABxωA+𝑜x=BxωxA+𝑜x=B
23 22 adantr AωBωABxωA+𝑜x=BxωxA+𝑜x=B
24 6 23 mpdd AωBωABxωxA+𝑜x=B
25 17 biimpa AωxωxAA+𝑜x
26 25 8 syl5ibcom AωxωxA+𝑜x=BAB
27 26 expimpd AωxωxA+𝑜x=BAB
28 27 rexlimdva AωxωxA+𝑜x=BAB
29 28 adantr AωBωxωxA+𝑜x=BAB
30 24 29 impbid AωBωABxωxA+𝑜x=B