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Theorem mul4r 11385
Description: Rearrangement of 4 factors: swap the right factors in the factors of a product of two products. (Contributed by AV, 4-Mar-2023.)
Assertion
Ref Expression
mul4r (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))

Proof of Theorem mul4r
StepHypRef Expression
1 mulcom 11192 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
21adantl 486 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
32oveq2d 7428 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐵) · (𝐷 · 𝐶)))
4 mul4 11384 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
54ancom2s 662 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
6 simplr 780 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐵 ∈ ℂ)
7 simprl 782 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐶 ∈ ℂ)
86, 7mulcomd 11236 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 · 𝐶) = (𝐶 · 𝐵))
98oveq2d 7428 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐷) · (𝐵 · 𝐶)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
103, 5, 93eqtrd 2801 1 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4  wa 400   = wceq 1569  wcel 2142  (class class class)co 7412  cc 11104   · cmul 11111
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-mulcl 11168  ax-mulcom 11170  ax-mulass 11172
This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-iota 6492  df-fv 6544  df-ov 7415
This theorem is used by:  bhmafibid1cn  15524  bhmafibid2cn  15525  2itscplem2  49587
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