Metamath Proof Explorer


Theorem nna0r

Description: Addition to zero. Remark in proof of Theorem 4K(2) of Enderton p. 81. Note: In this and later theorems, we deliberately avoid the more general ordinal versions of these theorems (in this case oa0r ) so that we can avoid ax-rep , which is not needed for finite recursive definitions. (Contributed by NM, 20-Sep-1995) (Revised by Mario Carneiro, 14-Nov-2014)

Ref Expression
Assertion nna0r Aω+𝑜A=A

Proof

Step Hyp Ref Expression
1 oveq2 x=+𝑜x=+𝑜
2 id x=x=
3 1 2 eqeq12d x=+𝑜x=x+𝑜=
4 oveq2 x=y+𝑜x=+𝑜y
5 id x=yx=y
6 4 5 eqeq12d x=y+𝑜x=x+𝑜y=y
7 oveq2 x=sucy+𝑜x=+𝑜sucy
8 id x=sucyx=sucy
9 7 8 eqeq12d x=sucy+𝑜x=x+𝑜sucy=sucy
10 oveq2 x=A+𝑜x=+𝑜A
11 id x=Ax=A
12 10 11 eqeq12d x=A+𝑜x=x+𝑜A=A
13 0elon On
14 oa0 On+𝑜=
15 13 14 ax-mp +𝑜=
16 peano1 ω
17 nnasuc ωyω+𝑜sucy=suc+𝑜y
18 16 17 mpan yω+𝑜sucy=suc+𝑜y
19 suceq +𝑜y=ysuc+𝑜y=sucy
20 19 eqeq2d +𝑜y=y+𝑜sucy=suc+𝑜y+𝑜sucy=sucy
21 18 20 syl5ibcom yω+𝑜y=y+𝑜sucy=sucy
22 3 6 9 12 15 21 finds Aω+𝑜A=A