Theorem metreslem 14020

Description: Lemma for metres 14023. (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 5122	. 2 ⊢ (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ (𝑅 × 𝑅))
2		ineq2 3332	. . . 4 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → ((𝑅 × 𝑅) ∩ dom 𝐷) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋)))
3		dmres 4930	. . . 4 ⊢ dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ dom 𝐷)
4		inxp 4763	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))
5		incom 3329	. . . . 5 ⊢ ((𝑋 × 𝑋) ∩ (𝑅 × 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
6	4, 5	eqtr3i 2200	. . . 4 ⊢ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)) = ((𝑅 × 𝑅) ∩ (𝑋 × 𝑋))
7	2, 3, 6	3eqtr4g 2235	. . 3 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → dom (𝐷 ↾ (𝑅 × 𝑅)) = ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅)))
8	7	reseq2d 4909	. 2 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ dom (𝐷 ↾ (𝑅 × 𝑅))) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))
9	1, 8	eqtr3id 2224	1 ⊢ (dom 𝐷 = (𝑋 × 𝑋) → (𝐷 ↾ (𝑅 × 𝑅)) = (𝐷 ↾ ((𝑋 ∩ 𝑅) × (𝑋 ∩ 𝑅))))

Syntax hints:   → wi 4   = wceq 1353   ∩ cin 3130   × cxp 4626  dom cdm 4628   ↾ cres 4630

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4123  ax-pow 4176  ax-pr 4211

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2741  df-un 3135  df-in 3137  df-ss 3144  df-pw 3579  df-sn 3600  df-pr 3601  df-op 3603  df-br 4006  df-opab 4067  df-xp 4634  df-rel 4635  df-cnv 4636  df-dm 4638  df-rn 4639  df-res 4640

This theorem is referenced by:  xmetres  14022  metres  14023