Theorem xrneq1 39103

Description: Equality theorem for the range Cartesian product. (Contributed by Peter Mazsa, 16-Dec-2020.)

Assertion

Ref	Expression
xrneq1	⊢ (𝐴 = 𝐵 → (𝐴 ⋉ 𝐶) = (𝐵 ⋉ 𝐶))

Proof of Theorem xrneq1

Step	Hyp	Ref	Expression
1		coeq2 5846	. . 3 ⊢ (𝐴 = 𝐵 → (◡(1st ↾ (V × V)) ∘ 𝐴) = (◡(1st ↾ (V × V)) ∘ 𝐵))
2	1	ineq1d 4172	. 2 ⊢ (𝐴 = 𝐵 → ((◡(1st ↾ (V × V)) ∘ 𝐴) ∩ (◡(2nd ↾ (V × V)) ∘ 𝐶)) = ((◡(1st ↾ (V × V)) ∘ 𝐵) ∩ (◡(2nd ↾ (V × V)) ∘ 𝐶)))
3		df-xrn 39087	. 2 ⊢ (𝐴 ⋉ 𝐶) = ((◡(1st ↾ (V × V)) ∘ 𝐴) ∩ (◡(2nd ↾ (V × V)) ∘ 𝐶))
4		df-xrn 39087	. 2 ⊢ (𝐵 ⋉ 𝐶) = ((◡(1st ↾ (V × V)) ∘ 𝐵) ∩ (◡(2nd ↾ (V × V)) ∘ 𝐶))
5	2, 3, 4	3eqtr4g 2825	1 ⊢ (𝐴 = 𝐵 → (𝐴 ⋉ 𝐶) = (𝐵 ⋉ 𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1570  Vcvv 3457   ∩ cin 3905   × cxp 5661  ^◡ccnv 5662   ↾ cres 5665   ∘ ccom 5667  1^st c1st 7990  2^nd c2nd 7991   ⋉ cxrn 38881

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-in 3913  df-ss 3923  df-br 5112  df-opab 5176  df-co 5672  df-xrn 39087

This theorem is used by:  xrneq1i  39104  xrneq1d  39105  xrneq12  39109