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Theorem mul4r 11324
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 11137 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
21adantl 482 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐶 · 𝐷) = (𝐷 · 𝐶))
32oveq2d 7373 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐶 · 𝐷)) = ((𝐴 · 𝐵) · (𝐷 · 𝐶)))
4 mul4 11323 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐷 ∈ ℂ ∧ 𝐶 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
54ancom2s 648 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐵) · (𝐷 · 𝐶)) = ((𝐴 · 𝐷) · (𝐵 · 𝐶)))
6 simplr 767 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐵 ∈ ℂ)
7 simprl 769 . . . 4 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → 𝐶 ∈ ℂ)
86, 7mulcomd 11176 . . 3 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → (𝐵 · 𝐶) = (𝐶 · 𝐵))
98oveq2d 7373 . 2 (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴 · 𝐷) · (𝐵 · 𝐶)) = ((𝐴 · 𝐷) · (𝐶 · 𝐵)))
103, 5, 93eqtrd 2780 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 396   = wceq 1541  wcel 2106  (class class class)co 7357  cc 11049   · cmul 11056
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2707  ax-mulcl 11113  ax-mulcom 11115  ax-mulass 11117
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-sb 2068  df-clab 2714  df-cleq 2728  df-clel 2814  df-rab 3408  df-v 3447  df-dif 3913  df-un 3915  df-in 3917  df-ss 3927  df-nul 4283  df-if 4487  df-sn 4587  df-pr 4589  df-op 4593  df-uni 4866  df-br 5106  df-iota 6448  df-fv 6504  df-ov 7360
This theorem is referenced by:  bhmafibid1cn  15348  bhmafibid2cn  15349  2itscplem2  46855
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