Theorem metreslem 13020

Description: Lemma for metres 13023. (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 5095	. 2 ⊢ (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ (𝑅 × 𝑅))
2		ineq2 3317	. . . 4 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → ((𝑅 × 𝑅) ∩ dom 𝐷) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋)))
3		dmres 4905	. . . 4 ⊢ dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ dom 𝐷)
4		inxp 4738	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))
5		incom 3314	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
6	4, 5	eqtr3i 2188	. . . 4 ⊢ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
7	2, 3, 6	3eqtr4g 2224	. . 3 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)))
8	7	reseq2d 4884	. 2 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))
9	1, 8	eqtr3id 2213	1 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ (𝑅 × 𝑅)) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))

Syntax hints:   → wi 4   = wceq 1343   ∩ cin 3115   × cxp 4602  dom cdm 4604   ↾ cres 4606

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-14 2139  ax-ext 2147  ax-sep 4100  ax-pow 4153  ax-pr 4187

This theorem depends on definitions:  df-bi 116  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-eu 2017  df-mo 2018  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-v 2728  df-un 3120  df-in 3122  df-ss 3129  df-pw 3561  df-sn 3582  df-pr 3583  df-op 3585  df-br 3983  df-opab 4044  df-xp 4610  df-rel 4611  df-cnv 4612  df-dm 4614  df-rn 4615  df-res 4616

This theorem is referenced by:  xmetres  13022  metres  13023