rrx2pnedifcoorneorr

Description: If two different points X and Y in a real Euclidean space of dimension 2 are different, then at least one difference of two corresponding coordinates is not 0. (Contributed by AV, 26-Feb-2023)

	Ref	Expression
Hypotheses	rrx2pnecoorneor.i	⊢ 𝐼 = { 1 , 2 }
	rrx2pnecoorneor.b	⊢ 𝑃 = ( ℝ ↑_m 𝐼 )
	rrx2pnedifcoorneor.a	⊢ 𝐴 = ( ( 𝑌 ‘ 1 ) − ( 𝑋 ‘ 1 ) )
	rrx2pnedifcoorneorr.b	⊢ 𝐵 = ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) )
Assertion	rrx2pnedifcoorneorr	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( 𝐴 ≠ 0 ∨ 𝐵 ≠ 0 ) )

Proof

Step	Hyp	Ref	Expression
1		rrx2pnecoorneor.i	⊢ 𝐼 = { 1 , 2 }
2		rrx2pnecoorneor.b	⊢ 𝑃 = ( ℝ ↑_m 𝐼 )
3		rrx2pnedifcoorneor.a	⊢ 𝐴 = ( ( 𝑌 ‘ 1 ) − ( 𝑋 ‘ 1 ) )
4		rrx2pnedifcoorneorr.b	⊢ 𝐵 = ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) )
5		eqid	⊢ ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) = ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) )
6	1 2 3 5	rrx2pnedifcoorneor	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( 𝐴 ≠ 0 ∨ ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) ≠ 0 ) )
7		eqcom	⊢ ( ( 𝑌 ‘ 2 ) = ( 𝑋 ‘ 2 ) ↔ ( 𝑋 ‘ 2 ) = ( 𝑌 ‘ 2 ) )
8	7	a1i	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( 𝑌 ‘ 2 ) = ( 𝑋 ‘ 2 ) ↔ ( 𝑋 ‘ 2 ) = ( 𝑌 ‘ 2 ) ) )
9	1 2	rrx2pyel	⊢ ( 𝑋 ∈ 𝑃 → ( 𝑋 ‘ 2 ) ∈ ℝ )
10	9	recnd	⊢ ( 𝑋 ∈ 𝑃 → ( 𝑋 ‘ 2 ) ∈ ℂ )
11	1 2	rrx2pyel	⊢ ( 𝑌 ∈ 𝑃 → ( 𝑌 ‘ 2 ) ∈ ℝ )
12	11	recnd	⊢ ( 𝑌 ∈ 𝑃 → ( 𝑌 ‘ 2 ) ∈ ℂ )
13	10 12	anim12i	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ) → ( ( 𝑋 ‘ 2 ) ∈ ℂ ∧ ( 𝑌 ‘ 2 ) ∈ ℂ ) )
14	13	ancomd	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ) → ( ( 𝑌 ‘ 2 ) ∈ ℂ ∧ ( 𝑋 ‘ 2 ) ∈ ℂ ) )
15	14	3adant3	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( 𝑌 ‘ 2 ) ∈ ℂ ∧ ( 𝑋 ‘ 2 ) ∈ ℂ ) )
16		subeq0	⊢ ( ( ( 𝑌 ‘ 2 ) ∈ ℂ ∧ ( 𝑋 ‘ 2 ) ∈ ℂ ) → ( ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) = 0 ↔ ( 𝑌 ‘ 2 ) = ( 𝑋 ‘ 2 ) ) )
17	15 16	syl	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) = 0 ↔ ( 𝑌 ‘ 2 ) = ( 𝑋 ‘ 2 ) ) )
18	13	3adant3	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( 𝑋 ‘ 2 ) ∈ ℂ ∧ ( 𝑌 ‘ 2 ) ∈ ℂ ) )
19		subeq0	⊢ ( ( ( 𝑋 ‘ 2 ) ∈ ℂ ∧ ( 𝑌 ‘ 2 ) ∈ ℂ ) → ( ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) ) = 0 ↔ ( 𝑋 ‘ 2 ) = ( 𝑌 ‘ 2 ) ) )
20	18 19	syl	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) ) = 0 ↔ ( 𝑋 ‘ 2 ) = ( 𝑌 ‘ 2 ) ) )
21	8 17 20	3bitr4d	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) = 0 ↔ ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) ) = 0 ) )
22	4	eqcomi	⊢ ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) ) = 𝐵
23	22	eqeq1i	⊢ ( ( ( 𝑋 ‘ 2 ) − ( 𝑌 ‘ 2 ) ) = 0 ↔ 𝐵 = 0 )
24	21 23	bitrdi	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) = 0 ↔ 𝐵 = 0 ) )
25	24	necon3bid	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) ≠ 0 ↔ 𝐵 ≠ 0 ) )
26	25	orbi2d	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( ( 𝐴 ≠ 0 ∨ ( ( 𝑌 ‘ 2 ) − ( 𝑋 ‘ 2 ) ) ≠ 0 ) ↔ ( 𝐴 ≠ 0 ∨ 𝐵 ≠ 0 ) ) )
27	6 26	mpbid	⊢ ( ( 𝑋 ∈ 𝑃 ∧ 𝑌 ∈ 𝑃 ∧ 𝑋 ≠ 𝑌 ) → ( 𝐴 ≠ 0 ∨ 𝐵 ≠ 0 ) )

Metamath Proof Explorer

Theorem rrx2pnedifcoorneorr

Proof