Theorem rspcsbnea 42848

Description: Special case related to rspsbc 3840. (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 3840	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → [𝐴 / 𝑥]𝐶 ≠ 𝐷))
2		df-ne 2966	. . . . . . 7 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
3	2	sbcbii 3808	. . . . . 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 3799	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ [𝐴 / 𝑥]𝐶 = 𝐷))
6		sbceq1g 4388	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
7	6	notbid 321	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → (¬ [𝐴 / 𝑥]𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
8	5, 7	bitrd 282	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
9	4, 8	bitrd 282	. . . 4 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
10		biidd 265	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → (⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
11	10	necon3bbid 3002	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
12	9, 11	bitrd 282	. . 3 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
13	1, 12	sylibd 242	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
14	13	imp 411	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1568   ∈ wcel 2150   ≠ wne 2965  ∀wral 3086  [wsbc 3752  ⦋csb 3861

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1823  ax-4 1837  ax-5 1938  ax-6 1995  ax-7 2036  ax-8 2152  ax-9 2160  ax-10 2183  ax-11 2199  ax-12 2220  ax-ext 2742

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1571  df-ex 1808  df-nf 1812  df-sb 2099  df-clab 2749  df-cleq 2762  df-clel 2845  df-nfc 2919  df-ne 2966  df-ral 3087  df-v 3464  df-sbc 3753  df-csb 3862

This theorem is referenced by:  idomnnzgmulnz  42850  deg1gprod  42857