Theorem rhmsubclem2 45645

Description: Lemma 2 for rhmsubc 45648. (Contributed by AV, 2-Mar-2020.)

Hypotheses

Ref	Expression
rngcrescrhm.u	⊢ (𝜑 → 𝑈 ∈ 𝑉)
rngcrescrhm.c	⊢ 𝐶 = (RngCat‘𝑈)
rngcrescrhm.r	⊢ (𝜑 → 𝑅 = (Ring ∩ 𝑈))
rngcrescrhm.h	⊢ 𝐻 = ( RingHom ↾ (𝑅 × 𝑅))

Assertion

Ref	Expression
rhmsubclem2	⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (𝑋𝐻𝑌) = (𝑋 RingHom 𝑌))

Proof of Theorem rhmsubclem2

Step	Hyp	Ref	Expression
1		opelxpi 5626	. . . 4 ⊢ ((𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → ⟨𝑋, 𝑌⟩ ∈ (𝑅 × 𝑅))
2	1	3adant1 1129	. . 3 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → ⟨𝑋, 𝑌⟩ ∈ (𝑅 × 𝑅))
3	2	fvresd 6794	. 2 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩) = ( RingHom ‘⟨𝑋, 𝑌⟩))
4		df-ov 7278	. . 3 ⊢ (𝑋𝐻𝑌) = (𝐻‘⟨𝑋, 𝑌⟩)
5		rngcrescrhm.h	. . . 4 ⊢ 𝐻 = ( RingHom ↾ (𝑅 × 𝑅))
6	5	fveq1i 6775	. . 3 ⊢ (𝐻‘⟨𝑋, 𝑌⟩) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
7	4, 6	eqtri 2766	. 2 ⊢ (𝑋𝐻𝑌) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
8		df-ov 7278	. 2 ⊢ (𝑋 RingHom 𝑌) = ( RingHom ‘⟨𝑋, 𝑌⟩)
9	3, 7, 8	3eqtr4g 2803	1 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (𝑋𝐻𝑌) = (𝑋 RingHom 𝑌))

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1539   ∈ wcel 2106   ∩ cin 3886  ⟨cop 4567   × cxp 5587   ↾ cres 5591  ‘cfv 6433  (class class class)co 7275  Ringcrg 19783   RingHom crh 19956  RngCatcrngc 45515

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-uni 4840  df-br 5075  df-opab 5137  df-xp 5595  df-res 5601  df-iota 6391  df-fv 6441  df-ov 7278

This theorem is referenced by:  rhmsubclem3  45646  rhmsubclem4  45647