Theorem metreslem 15094

Description: Lemma for metres 15097. (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 5226	. 2 ⊢ (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ (𝑅 × 𝑅))
2		ineq2 3400	. . . 4 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → ((𝑅 × 𝑅) ∩ dom 𝐷) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋)))
3		dmres 5032	. . . 4 ⊢ dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ dom 𝐷)
4		inxp 4862	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))
5		incom 3397	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
6	4, 5	eqtr3i 2252	. . . 4 ⊢ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
7	2, 3, 6	3eqtr4g 2287	. . 3 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)))
8	7	reseq2d 5011	. 2 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))
9	1, 8	eqtr3id 2276	1 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ (𝑅 × 𝑅)) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))

Syntax hints:   → wi 4   = wceq 1395   ∩ cin 3197   × cxp 4721  dom cdm 4723   ↾ cres 4725

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-14 2203  ax-ext 2211  ax-sep 4205  ax-pow 4262  ax-pr 4297

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-eu 2080  df-mo 2081  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2802  df-un 3202  df-in 3204  df-ss 3211  df-pw 3652  df-sn 3673  df-pr 3674  df-op 3676  df-br 4087  df-opab 4149  df-xp 4729  df-rel 4730  df-cnv 4731  df-dm 4733  df-rn 4734  df-res 4735

This theorem is referenced by:  xmetres  15096  metres  15097