Theorem rspcsbnea 42149

Description: Special case related to rspsbc 3859. (Contributed by metakunt, 5-May-2025.)

Assertion

Ref	Expression
rspcsbnea	⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

Distinct variable groups:   𝑥,𝐵   𝑥,𝐷

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

Proof of Theorem rspcsbnea

Step	Hyp	Ref	Expression
1		rspsbc 3859	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → [𝐴 / 𝑥]𝐶 ≠ 𝐷))
2		df-ne 2934	. . . . . . 7 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
3	2	sbcbii 3827	. . . . . 6 ⊢ ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ [𝐴 / 𝑥] ¬ 𝐶 = 𝐷)
4	3	a1i 11	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ [𝐴 / 𝑥] ¬ 𝐶 = 𝐷))
5		sbcng 3818	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ [𝐴 / 𝑥]𝐶 = 𝐷))
6		sbceq1g 4397	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
7	6	notbid 318	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → (¬ [𝐴 / 𝑥]𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
8	5, 7	bitrd 279	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
9	4, 8	bitrd 279	. . . 4 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
10		biidd 262	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → (⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
11	10	necon3bbid 2970	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
12	9, 11	bitrd 279	. . 3 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
13	1, 12	sylibd 239	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
14	13	imp 406	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540   ∈ wcel 2109   ≠ wne 2933  ∀wral 3052  [wsbc 3770  ⦋csb 3879

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 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-nfc 2886  df-ne 2934  df-ral 3053  df-v 3466  df-sbc 3771  df-csb 3880

This theorem is referenced by:  idomnnzgmulnz  42151  deg1gprod  42158