Theorem rhmsubclem2 20851

Description: Lemma 2 for rhmsubc 20854. (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 5692	. . . 4 ⊢ ((𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → ⟨𝑋, 𝑌⟩ ∈ (𝑅 × 𝑅))
2	1	3adant1 1148	. . 3 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → ⟨𝑋, 𝑌⟩ ∈ (𝑅 × 𝑅))
3	2	fvresd 6899	. 2 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩) = ( RingHom ‘⟨𝑋, 𝑌⟩))
4		df-ov 7417	. . 3 ⊢ (𝑋𝐻𝑌) = (𝐻‘⟨𝑋, 𝑌⟩)
5		rngcrescrhm.h	. . . 4 ⊢ 𝐻 = ( RingHom ↾ (𝑅 × 𝑅))
6	5	fveq1i 6880	. . 3 ⊢ (𝐻‘⟨𝑋, 𝑌⟩) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
7	4, 6	eqtri 2783	. 2 ⊢ (𝑋𝐻𝑌) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
8		df-ov 7417	. 2 ⊢ (𝑋 RingHom 𝑌) = ( RingHom ‘⟨𝑋, 𝑌⟩)
9	3, 7, 8	3eqtr4g 2820	1 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (𝑋𝐻𝑌) = (𝑋 RingHom 𝑌))

This proof depends on syntax axioms:   → wi 4   ∧ w3a 1103   = wceq 1570   ∈ wcel 2145   ∩ cin 3898  ⟨cop 4590   × cxp 5653   ↾ cres 5657  ‘cfv 6533  (class class class)co 7414  Ringcrg 20375   RingHom crh 20613  RngCatcrngc 20781

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732  ax-sep 5251  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-xp 5661  df-res 5667  df-iota 6489  df-fv 6541  df-ov 7417

This theorem is used by:  rhmsubclem3  20852  rhmsubclem4  20853