Theorem eqsbc2VD 46082

Description: Virtual deduction proof of eqsbc2 3805. (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 45817	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2		eqsbc1 3788	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
3	1, 2	e1a 45870	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶)   )
4		eqcom 2770	. . . . . . . . . 10 ⊢ (𝐶 = 𝑥 ↔ 𝑥 = 𝐶)
5	4	sbcbii 3798	. . . . . . . . 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 45870	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶)   )
8		idn2 45856	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   [𝐴 / 𝑥]𝐶 = 𝑥   )
9		biimp 218	. . . . . . 7 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶) → ([𝐴 / 𝑥]𝐶 = 𝑥 → [𝐴 / 𝑥]𝑥 = 𝐶))
10	7, 8, 9	e12 45966	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   [𝐴 / 𝑥]𝑥 = 𝐶   )
11		biimp 218	. . . . . 6 ⊢ (([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶) → ([𝐴 / 𝑥]𝑥 = 𝐶 → 𝐴 = 𝐶))
12	3, 10, 11	e12 45966	. . . . 5 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   𝐴 = 𝐶   )
13		eqcom 2770	. . . . 5 ⊢ (𝐴 = 𝐶 ↔ 𝐶 = 𝐴)
14	12, 13	e2bi 45875	. . . 4 ⊢ (   𝐴 ∈ 𝐵   ,   [𝐴 / 𝑥]𝐶 = 𝑥   ▶   𝐶 = 𝐴   )
15	14	in2 45848	. . 3 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐶 = 𝐴)   )
16		idn2 45856	. . . . . . 7 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   𝐶 = 𝐴   )
17	16, 13	e2bir 45876	. . . . . 6 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   𝐴 = 𝐶   )
18		biimpr 223	. . . . . 6 ⊢ (([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶) → (𝐴 = 𝐶 → [𝐴 / 𝑥]𝑥 = 𝐶))
19	3, 17, 18	e12 45966	. . . . 5 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   [𝐴 / 𝑥]𝑥 = 𝐶   )
20		biimpr 223	. . . . 5 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶) → ([𝐴 / 𝑥]𝑥 = 𝐶 → [𝐴 / 𝑥]𝐶 = 𝑥))
21	7, 19, 20	e12 45966	. . . 4 ⊢ (   𝐴 ∈ 𝐵   ,   𝐶 = 𝐴   ▶   [𝐴 / 𝑥]𝐶 = 𝑥   )
22	21	in2 45848	. . 3 ⊢ (   𝐴 ∈ 𝐵   ▶   (𝐶 = 𝐴 → [𝐴 / 𝑥]𝐶 = 𝑥)   )
23		impbi 211	. . 3 ⊢ (([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐶 = 𝐴) → ((𝐶 = 𝐴 → [𝐴 / 𝑥]𝐶 = 𝑥) → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴)))
24	15, 22, 23	e11 45931	. 2 ⊢ (   𝐴 ∈ 𝐵   ▶   ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴)   )
25	24	in1 45814	1 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1570   ∈ wcel 2147  [wsbc 3742

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-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-sbc 3743  df-vd1 45813  df-vd2 45821

This theorem is used by: (None)