Theorem rspcsbnea 43182

Description: Special case related to rspsbc 3825. (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 3825	. . 3 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷 → [𝐴 / 𝑥]𝐶 ≠ 𝐷))
2		df-ne 2956	. . . . . . 7 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
3	2	sbcbii 3794	. . . . . 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 3785	. . . . . 6 ⊢ (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥] ¬ 𝐶 = 𝐷 ↔ ¬ [𝐴 / 𝑥]𝐶 = 𝐷))
6		sbceq1g 4374	. . . . . . 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 2992	. . . 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 412	1 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥 ∈ 𝐵 𝐶 ≠ 𝐷) → ⦋𝐴 / 𝑥⦌𝐶 ≠ 𝐷)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144   ≠ wne 2955  ∀wral 3076  [wsbc 3738  ⦋csb 3846

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-v 3452  df-sbc 3739  df-csb 3847

This theorem is used by:  idomnnzgmulnz  43184  deg1gprod  43191