Theorem rspcsbnea 41634

Description: Special case related to rspsbc 3874. (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 3874	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → [𝐴 / 𝑥]𝐶 ≠ 𝐷))
2		df-ne 2938	. . . . . . 7 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
3	2	sbcbii 3839	. . . . . 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 3829	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ [𝐴 / 𝑥]𝐶 = 𝐷))
6		sbceq1g 4418	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
7	6	notbid 317	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → (¬ [𝐴 / 𝑥]𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
8	5, 7	bitrd 278	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
9	4, 8	bitrd 278	. . . 4 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
10		biidd 261	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → (⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
11	10	necon3bbid 2975	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
12	9, 11	bitrd 278	. . 3 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
13	1, 12	sylibd 238	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
14	13	imp 405	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 394   = wceq 1533   ∈ wcel 2098   ≠ wne 2937  ∀wral 3058  [wsbc 3778  ⦋csb 3894

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2166  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2706  df-cleq 2720  df-clel 2806  df-nfc 2881  df-ne 2938  df-ral 3059  df-v 3475  df-sbc 3779  df-csb 3895

This theorem is referenced by:  idomnnzgmulnz  41636  deg1gprod  41644