Theorem sbcpr 48330

Description: The proper substitution of an unordered pair for a setvar variable corresponds to a proper substitution of each of its elements. (Contributed by AV, 7-Apr-2023.)

Hypothesis

Ref	Expression
sbcpr.x	⊢ (𝑝 = {𝑥, 𝑦} → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
sbcpr	⊢ ([{𝑎, 𝑏} / 𝑝]𝜑 ↔ [𝑏 / 𝑦][𝑎 / 𝑥]𝜓)

Distinct variable groups:   𝑎,𝑏,𝑥,𝑦,𝑝   𝜑,𝑥,𝑦   𝜓,𝑝

Allowed substitution hints:   𝜑(𝑝, 𝑎, 𝑏)   𝜓(𝑥, 𝑦, 𝑎, 𝑏)

Proof of Theorem sbcpr

Step	Hyp	Ref	Expression
1		sbc5 3774	. . 3 ⊢ ([{𝑎, 𝑏} / 𝑝]𝜑 ↔ ∃𝑝(𝑝 = {𝑎, 𝑏} ∧ 𝜑))
2		preq12 4703	. . . . . . . . . . . . . . . . . 18 ⊢ ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → {𝑥, 𝑦} = {𝑎, 𝑏})
3	2	eqcomd 2771	. . . . . . . . . . . . . . . . 17 ⊢ ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → {𝑎, 𝑏} = {𝑥, 𝑦})
4	3	eqeq2d 2776	. . . . . . . . . . . . . . . 16 ⊢ ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → (𝑝 = {𝑎, 𝑏} ↔ 𝑝 = {𝑥, 𝑦}))
5	4	biimpa 482	. . . . . . . . . . . . . . 15 ⊢ (((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) ∧ 𝑝 = {𝑎, 𝑏}) → 𝑝 = {𝑥, 𝑦})
6		sbcpr.x	. . . . . . . . . . . . . . 15 ⊢ (𝑝 = {𝑥, 𝑦} → (𝜑 ↔ 𝜓))
7	5, 6	syl 18	. . . . . . . . . . . . . 14 ⊢ (((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) ∧ 𝑝 = {𝑎, 𝑏}) → (𝜑 ↔ 𝜓))
8	7	biimpd 232	. . . . . . . . . . . . 13 ⊢ (((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) ∧ 𝑝 = {𝑎, 𝑏}) → (𝜑 → 𝜓))
9	8	expcom 419	. . . . . . . . . . . 12 ⊢ (𝑝 = {𝑎, 𝑏} → ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → (𝜑 → 𝜓)))
10	9	expd 421	. . . . . . . . . . 11 ⊢ (𝑝 = {𝑎, 𝑏} → (𝑥 = 𝑎 → (𝑦 = 𝑏 → (𝜑 → 𝜓))))
11	10	com24 96	. . . . . . . . . 10 ⊢ (𝑝 = {𝑎, 𝑏} → (𝜑 → (𝑦 = 𝑏 → (𝑥 = 𝑎 → 𝜓))))
12	11	imp31 423	. . . . . . . . 9 ⊢ (((𝑝 = {𝑎, 𝑏} ∧ 𝜑) ∧ 𝑦 = 𝑏) → (𝑥 = 𝑎 → 𝜓))
13	12	alrimiv 1960	. . . . . . . 8 ⊢ (((𝑝 = {𝑎, 𝑏} ∧ 𝜑) ∧ 𝑦 = 𝑏) → ∀𝑥(𝑥 = 𝑎 → 𝜓))
14		vex 3461	. . . . . . . . 9 ⊢ 𝑎 ∈ V
15	14	sbc6 3777	. . . . . . . 8 ⊢ ([𝑎 / 𝑥]𝜓 ↔ ∀𝑥(𝑥 = 𝑎 → 𝜓))
16	13, 15	sylibr 237	. . . . . . 7 ⊢ (((𝑝 = {𝑎, 𝑏} ∧ 𝜑) ∧ 𝑦 = 𝑏) → [𝑎 / 𝑥]𝜓)
17	16	ex 418	. . . . . 6 ⊢ ((𝑝 = {𝑎, 𝑏} ∧ 𝜑) → (𝑦 = 𝑏 → [𝑎 / 𝑥]𝜓))
18	17	alrimiv 1960	. . . . 5 ⊢ ((𝑝 = {𝑎, 𝑏} ∧ 𝜑) → ∀𝑦(𝑦 = 𝑏 → [𝑎 / 𝑥]𝜓))
19		vex 3461	. . . . . 6 ⊢ 𝑏 ∈ V
20	19	sbc6 3777	. . . . 5 ⊢ ([𝑏 / 𝑦][𝑎 / 𝑥]𝜓 ↔ ∀𝑦(𝑦 = 𝑏 → [𝑎 / 𝑥]𝜓))
21	18, 20	sylibr 237	. . . 4 ⊢ ((𝑝 = {𝑎, 𝑏} ∧ 𝜑) → [𝑏 / 𝑦][𝑎 / 𝑥]𝜓)
22	21	exlimiv 1963	. . 3 ⊢ (∃𝑝(𝑝 = {𝑎, 𝑏} ∧ 𝜑) → [𝑏 / 𝑦][𝑎 / 𝑥]𝜓)
23	1, 22	sylbi 220	. 2 ⊢ ([{𝑎, 𝑏} / 𝑝]𝜑 → [𝑏 / 𝑦][𝑎 / 𝑥]𝜓)
24		sbc5 3774	. . . . 5 ⊢ ([𝑏 / 𝑦][𝑎 / 𝑥]𝜓 ↔ ∃𝑦(𝑦 = 𝑏 ∧ [𝑎 / 𝑥]𝜓))
25		sbc5 3774	. . . . . . . 8 ⊢ ([𝑎 / 𝑥]𝜓 ↔ ∃𝑥(𝑥 = 𝑎 ∧ 𝜓))
26	6	bicomd 226	. . . . . . . . . . . . . . 15 ⊢ (𝑝 = {𝑥, 𝑦} → (𝜓 ↔ 𝜑))
27	5, 26	syl 18	. . . . . . . . . . . . . 14 ⊢ (((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) ∧ 𝑝 = {𝑎, 𝑏}) → (𝜓 ↔ 𝜑))
28	27	biimpd 232	. . . . . . . . . . . . 13 ⊢ (((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) ∧ 𝑝 = {𝑎, 𝑏}) → (𝜓 → 𝜑))
29	28	expcom 419	. . . . . . . . . . . 12 ⊢ (𝑝 = {𝑎, 𝑏} → ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → (𝜓 → 𝜑)))
30	29	com13 89	. . . . . . . . . . 11 ⊢ (𝜓 → ((𝑥 = 𝑎 ∧ 𝑦 = 𝑏) → (𝑝 = {𝑎, 𝑏} → 𝜑)))
31	30	expd 421	. . . . . . . . . 10 ⊢ (𝜓 → (𝑥 = 𝑎 → (𝑦 = 𝑏 → (𝑝 = {𝑎, 𝑏} → 𝜑))))
32	31	impcom 413	. . . . . . . . 9 ⊢ ((𝑥 = 𝑎 ∧ 𝜓) → (𝑦 = 𝑏 → (𝑝 = {𝑎, 𝑏} → 𝜑)))
33	32	exlimiv 1963	. . . . . . . 8 ⊢ (∃𝑥(𝑥 = 𝑎 ∧ 𝜓) → (𝑦 = 𝑏 → (𝑝 = {𝑎, 𝑏} → 𝜑)))
34	25, 33	sylbi 220	. . . . . . 7 ⊢ ([𝑎 / 𝑥]𝜓 → (𝑦 = 𝑏 → (𝑝 = {𝑎, 𝑏} → 𝜑)))
35	34	impcom 413	. . . . . 6 ⊢ ((𝑦 = 𝑏 ∧ [𝑎 / 𝑥]𝜓) → (𝑝 = {𝑎, 𝑏} → 𝜑))
36	35	exlimiv 1963	. . . . 5 ⊢ (∃𝑦(𝑦 = 𝑏 ∧ [𝑎 / 𝑥]𝜓) → (𝑝 = {𝑎, 𝑏} → 𝜑))
37	24, 36	sylbi 220	. . . 4 ⊢ ([𝑏 / 𝑦][𝑎 / 𝑥]𝜓 → (𝑝 = {𝑎, 𝑏} → 𝜑))
38	37	alrimiv 1960	. . 3 ⊢ ([𝑏 / 𝑦][𝑎 / 𝑥]𝜓 → ∀𝑝(𝑝 = {𝑎, 𝑏} → 𝜑))
39		prex 5411	. . . 4 ⊢ {𝑎, 𝑏} ∈ V
40	39	sbc6 3777	. . 3 ⊢ ([{𝑎, 𝑏} / 𝑝]𝜑 ↔ ∀𝑝(𝑝 = {𝑎, 𝑏} → 𝜑))
41	38, 40	sylibr 237	. 2 ⊢ ([𝑏 / 𝑦][𝑎 / 𝑥]𝜓 → [{𝑎, 𝑏} / 𝑝]𝜑)
42	23, 41	impbii 212	1 ⊢ ([{𝑎, 𝑏} / 𝑝]𝜑 ↔ [𝑏 / 𝑦][𝑎 / 𝑥]𝜓)

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812  [wsbc 3746  {cpr 4593

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-10 2179  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-pr 5406

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-sbc 3747  df-un 3911  df-sn 4592  df-pr 4594

This theorem is used by:  reupr  48331