Theorem addcomnq

Description: Addition of positive fractions is commutative. (Contributed by NM, 30-Aug-1995) (Revised by Mario Carneiro, 28-Apr-2013) (New usage is discouraged.)

	Ref	Expression
Assertion	addcomnq	$$⊢A{+}_{𝑸}B=B{+}_{𝑸}A$$

Proof

Step	Hyp	Ref	Expression
1		addcompq	$$⊢A{+}_{\mathrm{𝑝𝑸}}B=B{+}_{\mathrm{𝑝𝑸}}A$$
2	1	fveq2i	$$⊢{/}_{𝑸}\left(A{+}_{\mathrm{𝑝𝑸}}B\right)={/}_{𝑸}\left(B{+}_{\mathrm{𝑝𝑸}}A\right)$$
3		addpqnq	$$⊢\left(A\in 𝑸\wedge B\in 𝑸\right)\to A{+}_{𝑸}B={/}_{𝑸}\left(A{+}_{\mathrm{𝑝𝑸}}B\right)$$
4		addpqnq	$$⊢\left(B\in 𝑸\wedge A\in 𝑸\right)\to B{+}_{𝑸}A={/}_{𝑸}\left(B{+}_{\mathrm{𝑝𝑸}}A\right)$$
5	4	ancoms	$$⊢\left(A\in 𝑸\wedge B\in 𝑸\right)\to B{+}_{𝑸}A={/}_{𝑸}\left(B{+}_{\mathrm{𝑝𝑸}}A\right)$$
6	2 3 5	3eqtr4a	$$⊢\left(A\in 𝑸\wedge B\in 𝑸\right)\to A{+}_{𝑸}B=B{+}_{𝑸}A$$
7		addnqf	$$⊢{+}_{𝑸}:𝑸\times 𝑸⟶𝑸$$
8	7	fdmi	$$⊢\mathrm{dom}{+}_{𝑸}=𝑸\times 𝑸$$
9	8	ndmovcom	$$⊢\neg \left(A\in 𝑸\wedge B\in 𝑸\right)\to A{+}_{𝑸}B=B{+}_{𝑸}A$$
10	6 9	pm2.61i	$$⊢A{+}_{𝑸}B=B{+}_{𝑸}A$$