Theorem mulsval2lem 28334

Description: Lemma for mulsval2 28335. Change bound variables in one of the cases. (Contributed by Scott Fenton, 8-Mar-2025.)

Assertion

Ref	Expression
mulsval2lem	⊢ {𝑎 ∣ ∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} = {𝑏 ∣ ∃𝑟 ∈ 𝑋 ∃𝑠 ∈ 𝑌 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}

Distinct variable groups:   𝐴,𝑎,𝑏,𝑝,𝑞,𝑟   𝐴,𝑠   𝐵,𝑎,𝑏,𝑝,𝑞,𝑟   𝐵,𝑠   𝑋,𝑎,𝑏,𝑝,𝑟   𝑌,𝑎,𝑏,𝑝,𝑞,𝑟   𝑌,𝑠,𝑎,𝑏,𝑞,𝑟

Allowed substitution hints:   𝑋(𝑠, 𝑞)

Proof of Theorem mulsval2lem

Step	Hyp	Ref	Expression
1		eqeq1 2769	. . . 4 ⊢ (𝑎 = 𝑏 → (𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) ↔ 𝑏 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))))
2	1	2rexbidv 3232	. . 3 ⊢ (𝑎 = 𝑏 → (∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) ↔ ∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑏 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))))
3		oveq1 7423	. . . . . . 7 ⊢ (𝑝 = 𝑟 → (𝑝 ·s 𝐵) = (𝑟 ·s 𝐵))
4	3	oveq1d 7431	. . . . . 6 ⊢ (𝑝 = 𝑟 → ((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) = ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)))
5		oveq1 7423	. . . . . 6 ⊢ (𝑝 = 𝑟 → (𝑝 ·s 𝑞) = (𝑟 ·s 𝑞))
6	4, 5	oveq12d 7434	. . . . 5 ⊢ (𝑝 = 𝑟 → (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑟 ·s 𝑞)))
7	6	eqeq2d 2776	. . . 4 ⊢ (𝑝 = 𝑟 → (𝑏 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) ↔ 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑟 ·s 𝑞))))
8		oveq2 7424	. . . . . . 7 ⊢ (𝑞 = 𝑠 → (𝐴 ·s 𝑞) = (𝐴 ·s 𝑠))
9	8	oveq2d 7432	. . . . . 6 ⊢ (𝑞 = 𝑠 → ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) = ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)))
10		oveq2 7424	. . . . . 6 ⊢ (𝑞 = 𝑠 → (𝑟 ·s 𝑞) = (𝑟 ·s 𝑠))
11	9, 10	oveq12d 7434	. . . . 5 ⊢ (𝑞 = 𝑠 → (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑟 ·s 𝑞)) = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠)))
12	11	eqeq2d 2776	. . . 4 ⊢ (𝑞 = 𝑠 → (𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑟 ·s 𝑞)) ↔ 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))))
13	7, 12	cbvrex2vw 3250	. . 3 ⊢ (∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑏 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) ↔ ∃𝑟 ∈ 𝑋 ∃𝑠 ∈ 𝑌 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠)))
14	2, 13	bitrdi 290	. 2 ⊢ (𝑎 = 𝑏 → (∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞)) ↔ ∃𝑟 ∈ 𝑋 ∃𝑠 ∈ 𝑌 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))))
15	14	cbvabv 2835	1 ⊢ {𝑎 ∣ ∃𝑝 ∈ 𝑋 ∃𝑞 ∈ 𝑌 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} = {𝑏 ∣ ∃𝑟 ∈ 𝑋 ∃𝑠 ∈ 𝑌 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}

This proof depends on syntax axioms:   = wceq 1570  {cab 2743  ∃wrex 3091  (class class class)co 7416   +_s cadds 28183   -_s csubs 28244   ·_s cmuls 28330

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-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-br 5112  df-iota 6496  df-fv 6548  df-ov 7419

This theorem is used by:  mulsval2  28335  mulcut  28356  mulsunif  28374