Theorem rhmsubclem2 20953

Description: Lemma 2 for rhmsubc 20956. (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 6916	. 2 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩) = ( RingHom ‘⟨𝑋, 𝑌⟩))
4		df-ov 7433	. . 3 ⊢ (𝑋𝐻𝑌) = (𝐻‘⟨𝑋, 𝑌⟩)
5		rngcrescrhm.h	. . . 4 ⊢ 𝐻 = ( RingHom ↾ (𝑅 × 𝑅))
6	5	fveq1i 6897	. . 3 ⊢ (𝐻‘⟨𝑋, 𝑌⟩) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
7	4, 6	eqtri 2786	. 2 ⊢ (𝑋𝐻𝑌) = (( RingHom ↾ (𝑅 × 𝑅))‘⟨𝑋, 𝑌⟩)
8		df-ov 7433	. 2 ⊢ (𝑋 RingHom 𝑌) = ( RingHom ‘⟨𝑋, 𝑌⟩)
9	3, 7, 8	3eqtr4g 2823	1 ⊢ ((𝜑 ∧ 𝑋 ∈ 𝑅 ∧ 𝑌 ∈ 𝑅) → (𝑋𝐻𝑌) = (𝑋 RingHom 𝑌))

This proof depends on syntax axioms:   → wi 4   ∧ w3a 1103   = wceq 1569   ∈ wcel 2146   ∩ cin 4092  ⟨cop 4838   × cxp 5653   ↾ cres 5657  ‘cfv 6547  (class class class)co 7430  Ringcrg 20474   RingHom crh 20714  RngCatcrngc 20883

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735  ax-sep 5252  ax-pr 5395

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5171  df-xp 5661  df-res 5667  df-iota 6501  df-fv 6555  df-ov 7433

This theorem is used by:  rhmsubclem3  20954  rhmsubclem4  20955