Theorem eqsbc3r 2819

Description: eqsbc3 2802 with setvar variable on right side of equals sign. (Contributed by Alan Sare, 24-Oct-2011.)

Assertion

Ref	Expression
eqsbc3r	⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴))

Distinct variable groups:   𝑥,𝐶   𝑥,𝐴

Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem eqsbc3r

Step	Hyp	Ref	Expression
1		eqcom 2042	. . . . . 6 ⊢ (𝐶 = 𝑥 ↔ 𝑥 = 𝐶)
2	1	sbcbii 2818	. . . . 5 ⊢ ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ [𝐴 / 𝑥]𝑥 = 𝐶)
3	2	biimpi 113	. . . 4 ⊢ ([𝐴 / 𝑥]𝐶 = 𝑥 → [𝐴 / 𝑥]𝑥 = 𝐶)
4		eqsbc3 2802	. . . 4 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝑥 = 𝐶 ↔ 𝐴 = 𝐶))
5	3, 4	syl5ib 143	. . 3 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐴 = 𝐶))
6		eqcom 2042	. . 3 ⊢ (𝐴 = 𝐶 ↔ 𝐶 = 𝐴)
7	5, 6	syl6ib 150	. 2 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 → 𝐶 = 𝐴))
8		idd 21	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → (𝐶 = 𝐴 → 𝐶 = 𝐴))
9	8, 6	syl6ibr 151	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (𝐶 = 𝐴 → 𝐴 = 𝐶))
10	9, 4	sylibrd 158	. . 3 ⊢ (𝐴 ∈ 𝐵 → (𝐶 = 𝐴 → [𝐴 / 𝑥]𝑥 = 𝐶))
11	10, 2	syl6ibr 151	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐶 = 𝐴 → [𝐴 / 𝑥]𝐶 = 𝑥))
12	7, 11	impbid 120	1 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝑥 ↔ 𝐶 = 𝐴))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243   ∈ wcel 1393  [wsbc 2764

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sbc 2765

This theorem is referenced by: (None)