Theorem mulsval2lem 28553

Description: Lemma for mulsval2 28554. 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 2767	. . . 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 3231	. . 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 7458	. . . . . . 7 ⊢ (𝑝 = 𝑟 → (𝑝 ·s 𝐵) = (𝑟 ·s 𝐵))
4	3	oveq1d 7466	. . . . . 6 ⊢ (𝑝 = 𝑟 → ((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) = ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)))
5		oveq1 7458	. . . . . 6 ⊢ (𝑝 = 𝑟 → (𝑝 ·s 𝑞) = (𝑟 ·s 𝑞))
6	4, 5	oveq12d 7469	. . . . 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 2774	. . . 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 7459	. . . . . . 7 ⊢ (𝑞 = 𝑠 → (𝐴 ·s 𝑞) = (𝐴 ·s 𝑠))
9	8	oveq2d 7467	. . . . . 6 ⊢ (𝑞 = 𝑠 → ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑞)) = ((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)))
10		oveq2 7459	. . . . . 6 ⊢ (𝑞 = 𝑠 → (𝑟 ·s 𝑞) = (𝑟 ·s 𝑠))
11	9, 10	oveq12d 7469	. . . . 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 2774	. . . 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 3249	. . 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 2834	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 2741  ∃wrex 3090  (class class class)co 7451   +_s cadds 28402   −_s csubs 28463   ·_s cmuls 28549

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-iota 6520  df-fv 6574  df-ov 7454

This theorem is used by:  mulsval2  28554  mulcut  28575  mulsunif  28593