Metamath Proof Explorer


Theorem naddoa

Description: Natural addition of a natural is the same as regular addition. (Contributed by Scott Fenton, 20-Aug-2025)

Ref Expression
Assertion naddoa A On B ω A + B = A + 𝑜 B

Proof

Step Hyp Ref Expression
1 oveq2 y = A + y = A +
2 oveq2 y = A + 𝑜 y = A + 𝑜
3 1 2 eqeq12d y = A + y = A + 𝑜 y A + = A + 𝑜
4 3 imbi2d y = A On A + y = A + 𝑜 y A On A + = A + 𝑜
5 oveq2 y = x A + y = A + x
6 oveq2 y = x A + 𝑜 y = A + 𝑜 x
7 5 6 eqeq12d y = x A + y = A + 𝑜 y A + x = A + 𝑜 x
8 7 imbi2d y = x A On A + y = A + 𝑜 y A On A + x = A + 𝑜 x
9 oveq2 y = suc x A + y = A + suc x
10 oveq2 y = suc x A + 𝑜 y = A + 𝑜 suc x
11 9 10 eqeq12d y = suc x A + y = A + 𝑜 y A + suc x = A + 𝑜 suc x
12 11 imbi2d y = suc x A On A + y = A + 𝑜 y A On A + suc x = A + 𝑜 suc x
13 oveq2 y = B A + y = A + B
14 oveq2 y = B A + 𝑜 y = A + 𝑜 B
15 13 14 eqeq12d y = B A + y = A + 𝑜 y A + B = A + 𝑜 B
16 15 imbi2d y = B A On A + y = A + 𝑜 y A On A + B = A + 𝑜 B
17 naddrid A On A + = A
18 oa0 A On A + 𝑜 = A
19 17 18 eqtr4d A On A + = A + 𝑜
20 suceq A + x = A + 𝑜 x suc A + x = suc A + 𝑜 x
21 20 3ad2ant3 x ω A On A + x = A + 𝑜 x suc A + x = suc A + 𝑜 x
22 nnon x ω x On
23 naddsuc2 A On x On A + suc x = suc A + x
24 22 23 sylan2 A On x ω A + suc x = suc A + x
25 24 ancoms x ω A On A + suc x = suc A + x
26 25 3adant3 x ω A On A + x = A + 𝑜 x A + suc x = suc A + x
27 onasuc A On x ω A + 𝑜 suc x = suc A + 𝑜 x
28 27 ancoms x ω A On A + 𝑜 suc x = suc A + 𝑜 x
29 28 3adant3 x ω A On A + x = A + 𝑜 x A + 𝑜 suc x = suc A + 𝑜 x
30 21 26 29 3eqtr4d x ω A On A + x = A + 𝑜 x A + suc x = A + 𝑜 suc x
31 30 3exp x ω A On A + x = A + 𝑜 x A + suc x = A + 𝑜 suc x
32 31 a2d x ω A On A + x = A + 𝑜 x A On A + suc x = A + 𝑜 suc x
33 4 8 12 16 19 32 finds B ω A On A + B = A + 𝑜 B
34 33 impcom A On B ω A + B = A + 𝑜 B