Theorem eqsbc2VD 44864

Description: Virtual deduction proof of eqsbc2 3829. (Contributed by Alan Sare, 24-Oct-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
eqsbc2VD	⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴))

Distinct variable group:   𝑥,𝐶

Allowed substitution hints:   𝐴(𝑥)   𝐵(𝑥)

Proof of Theorem eqsbc2VD

Step	Hyp	Ref	Expression
1		idn1 44599	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2		eqsbc1 3812	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	e1a 44652	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶)   )
4		eqcom 2742	. . . . . . . . . 10 ⊢ (𝐶 = 𝑥 ↔ 𝑥 = 𝐶)
5	4	sbcbii 3822	. . . . . . . . 9 ⊢ ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶)
6	5	a1i 11	. . . . . . . 8 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶))
7	1, 6	e1a 44652	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶)   )
8		idn2 44638	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   [𝐴 / 𝑥]𝐶 = 𝑥   )
9		biimp 215	. . . . . . 7 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶) → ([𝐴 / 𝑥]𝐶 = 𝑥 → [𝐴 / 𝑥]𝑥 = 𝐶))
10	7, 8, 9	e12 44748	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   [𝐴 / 𝑥]𝑥 = 𝐶   )
11		biimp 215	. . . . . 6 ⊢ (([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶) → ([𝐴 / 𝑥]𝑥 = 𝐶 → 𝐴 = 𝐶))
12	3, 10, 11	e12 44748	. . . . 5 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   𝐴 = 𝐶   )
13		eqcom 2742	. . . . 5 ⊢ (𝐴 = 𝐶 ↔ 𝐶 = 𝐴)
14	12, 13	e2bi 44657	. . . 4 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   𝐶 = 𝐴   )
15	14	in2 44630	. . 3 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐶 = 𝐴)   )
16		idn2 44638	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   𝐶 = 𝐴   )
17	16, 13	e2bir 44658	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   𝐴 = 𝐶   )
18		biimpr 220	. . . . . 6 ⊢ (([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶) → (𝐴 = 𝐶 → [𝐴 / 𝑥]𝑥 = 𝐶))
19	3, 17, 18	e12 44748	. . . . 5 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   [𝐴 / 𝑥]𝑥 = 𝐶   )
20		biimpr 220	. . . . 5 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶) → ([𝐴 / 𝑥]𝑥 = 𝐶 → [𝐴 / 𝑥]𝐶 = 𝑥))
21	7, 19, 20	e12 44748	. . . 4 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   [𝐴 / 𝑥]𝐶 = 𝑥   )
22	21	in2 44630	. . 3 ⊢ (   𝐴 ∈ 𝐵   ▶   (𝐶 = 𝐴 → [𝐴 / 𝑥]𝐶 = 𝑥)   )
23		impbi 208	. . 3 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐶 = 𝐴) → ((𝐶 = 𝐴 → [𝐴 / 𝑥]𝐶 = 𝑥) → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴)))
24	15, 22, 23	e11 44713	. 2 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴)   )
25	24	in1 44596	1 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1540   ∈ wcel 2108  [wsbc 3765

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-sbc 3766  df-vd1 44595  df-vd2 44603

This theorem is referenced by: (None)