Theorem rspc6v 3602

Description: 6-variable restricted specialization, using implicit substitution. (Contributed by Scott Fenton, 20-Feb-2025.)

Hypotheses

Ref	Expression
rspc6v.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜒))
rspc6v.2	⊢ (𝑦 = 𝐵 → (𝜒 ↔ 𝜃))
rspc6v.3	⊢ (𝑧 = 𝐶 → (𝜃 ↔ 𝜏))
rspc6v.4	⊢ (𝑤 = 𝐷 → (𝜏 ↔ 𝜂))
rspc6v.5	⊢ (𝑝 = 𝐸 → (𝜂 ↔ 𝜁))
rspc6v.6	⊢ (𝑞 = 𝐹 → (𝜁 ↔ 𝜓))

Assertion

Ref	Expression
rspc6v	⊢ (((𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) ∧ (𝐶 ∈ 𝑇 ∧ 𝐷 ∈ 𝑈) ∧ (𝐸 ∈ 𝑉 ∧ 𝐹 ∈ 𝑊)) → (∀𝑥 ∈ 𝑅 ∀𝑦 ∈ 𝑆 ∀𝑧 ∈ 𝑇 ∀𝑤 ∈ 𝑈 ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜑 → 𝜓))

Distinct variable groups:   𝑥,𝐴,𝑦,𝑧,𝑤,𝑝,𝑞   𝑦,𝐵,𝑧,𝑤,𝑝,𝑞   𝑧,𝐶,𝑤,𝑝,𝑞   𝑤,𝐷,𝑝,𝑞   𝐸,𝑝,𝑞   𝐹,𝑞   𝑥,𝑅   𝑥,𝑆,𝑦   𝑥,𝑇,𝑦,𝑧   𝑥,𝑈,𝑦,𝑧,𝑤   𝑥,𝑉,𝑦,𝑧,𝑤,𝑝   𝑥,𝑊,𝑦,𝑧,𝑤,𝑝,𝑞   𝜒,𝑥   𝜃,𝑦   𝜏,𝑧   𝜂,𝑤   𝜁,𝑝   𝜓,𝑞

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧, 𝑤, 𝑞, 𝑝)   𝜓(𝑥, 𝑦, 𝑧, 𝑤, 𝑝)   𝜒(𝑦, 𝑧, 𝑤, 𝑞, 𝑝)   𝜃(𝑥, 𝑧, 𝑤, 𝑞, 𝑝)   𝜏(𝑥, 𝑦, 𝑤, 𝑞, 𝑝)   𝜂(𝑥, 𝑦, 𝑧, 𝑞, 𝑝)   𝜁(𝑥, 𝑦, 𝑧, 𝑤, 𝑞)   𝐵(𝑥)   𝐶(𝑥, 𝑦)   𝐷(𝑥, 𝑦, 𝑧)   𝑅(𝑦, 𝑧, 𝑤, 𝑞, 𝑝)   𝑆(𝑧, 𝑤, 𝑞, 𝑝)   𝑇(𝑤, 𝑞, 𝑝)   𝑈(𝑞, 𝑝)   𝐸(𝑥, 𝑦, 𝑧, 𝑤)   𝐹(𝑥, 𝑦, 𝑧, 𝑤, 𝑝)   𝑉(𝑞)

Proof of Theorem rspc6v

Step	Hyp	Ref	Expression
1		rspc6v.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜒))
2	1	2ralbidv 3229	. . . 4 ⊢ (𝑥 = 𝐴 → (∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜑 ↔ ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜒))
3		rspc6v.2	. . . . 5 ⊢ (𝑦 = 𝐵 → (𝜒 ↔ 𝜃))
4	3	2ralbidv 3229	. . . 4 ⊢ (𝑦 = 𝐵 → (∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜒 ↔ ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜃))
5		rspc6v.3	. . . . 5 ⊢ (𝑧 = 𝐶 → (𝜃 ↔ 𝜏))
6	5	2ralbidv 3229	. . . 4 ⊢ (𝑧 = 𝐶 → (∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜃 ↔ ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜏))
7		rspc6v.4	. . . . 5 ⊢ (𝑤 = 𝐷 → (𝜏 ↔ 𝜂))
8	7	2ralbidv 3229	. . . 4 ⊢ (𝑤 = 𝐷 → (∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜏 ↔ ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜂))
9	2, 4, 6, 8	rspc4v 3601	. . 3 ⊢ (((𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) ∧ (𝐶 ∈ 𝑇 ∧ 𝐷 ∈ 𝑈)) → (∀𝑥 ∈ 𝑅 ∀𝑦 ∈ 𝑆 ∀𝑧 ∈ 𝑇 ∀𝑤 ∈ 𝑈 ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜑 → ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜂))
10		rspc6v.5	. . . 4 ⊢ (𝑝 = 𝐸 → (𝜂 ↔ 𝜁))
11		rspc6v.6	. . . 4 ⊢ (𝑞 = 𝐹 → (𝜁 ↔ 𝜓))
12	10, 11	rspc2v 3592	. . 3 ⊢ ((𝐸 ∈ 𝑉 ∧ 𝐹 ∈ 𝑊) → (∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜂 → 𝜓))
13	9, 12	syl9 78	. 2 ⊢ (((𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) ∧ (𝐶 ∈ 𝑇 ∧ 𝐷 ∈ 𝑈)) → ((𝐸 ∈ 𝑉 ∧ 𝐹 ∈ 𝑊) → (∀𝑥 ∈ 𝑅 ∀𝑦 ∈ 𝑆 ∀𝑧 ∈ 𝑇 ∀𝑤 ∈ 𝑈 ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜑 → 𝜓)))
14	13	3impia 1135	1 ⊢ (((𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) ∧ (𝐶 ∈ 𝑇 ∧ 𝐷 ∈ 𝑈) ∧ (𝐸 ∈ 𝑉 ∧ 𝐹 ∈ 𝑊)) → (∀𝑥 ∈ 𝑅 ∀𝑦 ∈ 𝑆 ∀𝑧 ∈ 𝑇 ∀𝑤 ∈ 𝑈 ∀𝑝 ∈ 𝑉 ∀𝑞 ∈ 𝑊 𝜑 → 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 400   ∧ w3a 1103   = wceq 1570   ∈ wcel 2143  ∀wral 3079

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 401  df-3an 1105  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080

This theorem is used by:  mulsproplem1  28318  mulsprop  28332