Theorem metreslem 15103

Description: Lemma for metres 15106. (Contributed by Mario Carneiro, 24-Aug-2015.)

Assertion

Ref	Expression
metreslem	⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ (𝑅 × 𝑅)) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))

Proof of Theorem metreslem

Step	Hyp	Ref	Expression
1		resdmres 5228	. 2 ⊢ (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ (𝑅 × 𝑅))
2		ineq2 3402	. . . 4 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → ((𝑅 × 𝑅) ∩ dom 𝐷) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋)))
3		dmres 5034	. . . 4 ⊢ dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ dom 𝐷)
4		inxp 4864	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))
5		incom 3399	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
6	4, 5	eqtr3i 2254	. . . 4 ⊢ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
7	2, 3, 6	3eqtr4g 2289	. . 3 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)))
8	7	reseq2d 5013	. 2 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))
9	1, 8	eqtr3id 2278	1 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ (𝑅 × 𝑅)) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))

Syntax hints:   → wi 4   = wceq 1397   ∩ cin 3199   × cxp 4723  dom cdm 4725   ↾ cres 4727

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-14 2205  ax-ext 2213  ax-sep 4207  ax-pow 4264  ax-pr 4299

This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-eu 2082  df-mo 2083  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-v 2804  df-un 3204  df-in 3206  df-ss 3213  df-pw 3654  df-sn 3675  df-pr 3676  df-op 3678  df-br 4089  df-opab 4151  df-xp 4731  df-rel 4732  df-cnv 4733  df-dm 4735  df-rn 4736  df-res 4737

This theorem is referenced by:  xmetres  15105  metres  15106