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Theorem mul4r 11288
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 11098 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
21adantl 481 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
32oveq2d 7368 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐵) · (𝐷 · 𝐶)))
4 mul4 11287 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
54ancom2s 650 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
6 simplr 768 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐵 ∈ ℂ)
7 simprl 770 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐶 ∈ ℂ)
86, 7mulcomd 11139 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 · 𝐶) = (𝐶 · 𝐵))
98oveq2d 7368 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐷) · (𝐵 · 𝐶)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
103, 5, 93eqtrd 2770 1 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1541  wcel 2111  (class class class)co 7352  cc 11010   · cmul 11017
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703  ax-mulcl 11074  ax-mulcom 11076  ax-mulass 11078
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-dif 3900  df-un 3902  df-ss 3914  df-nul 4283  df-if 4475  df-sn 4576  df-pr 4578  df-op 4582  df-uni 4859  df-br 5094  df-iota 6443  df-fv 6495  df-ov 7355
This theorem is referenced by:  bhmafibid1cn  15379  bhmafibid2cn  15380  2itscplem2  48885
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