Theorem rspcsbnea 42753

Description: Special case related to rspsbc 3834. (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 3834	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → [𝐴 / 𝑥]𝐶 ≠ 𝐷))
2		df-ne 2960	. . . . . . 7 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
3	2	sbcbii 3802	. . . . . 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 3793	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ [𝐴 / 𝑥]𝐶 = 𝐷))
6		sbceq1g 4373	. . . . . . 7 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
7	6	notbid 320	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → (¬ [𝐴 / 𝑥]𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
8	5, 7	bitrd 281	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
9	4, 8	bitrd 281	. . . 4 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
10		biidd 264	. . . . 5 ⊢ (𝐴 ∈ 𝐵 → (⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷))
11	10	necon3bbid 2996	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (¬ ⦋𝐴 / 𝑥⦌𝐶 = 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
12	9, 11	bitrd 281	. . 3 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥]𝐶 ≠ 𝐷 ↔ ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
13	1, 12	sylibd 241	. 2 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷))
14	13	imp 410	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1562   ∈ wcel 2144   ≠ wne 2959  ∀wral 3078  [wsbc 3746  ⦋csb 3854

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2214  ax-ext 2736

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1565  df-ex 1802  df-nf 1806  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3079  df-v 3458  df-sbc 3747  df-csb 3855

This theorem is referenced by:  idomnnzgmulnz  42755  deg1gprod  42762