Theorem sbcne12g 3119

Description: Distribute proper substitution through an inequality. (Contributed by Andrew Salmon, 18-Jun-2011.)

Assertion

Ref	Expression
sbcne12g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶))

Proof of Theorem sbcne12g

Step	Hyp	Ref	Expression
1		sbceqg 3117	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2	1	notbid 669	. 2 ⊢ (𝐴 ∈ 𝑉 → (¬ [𝐴 / 𝑥]𝐵 = 𝐶 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
3		df-ne 2379	. . . . 5 ⊢ (𝐵 ≠ 𝐶 ↔ ¬ 𝐵 = 𝐶)
4	3	sbcbii 3065	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ [𝐴 / 𝑥] ¬ 𝐵 = 𝐶)
5		sbcng 3046	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥] ¬ 𝐵 = 𝐶 ↔ ¬ [𝐴 / 𝑥]𝐵 = 𝐶))
6	4, 5	bitrid 192	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ ¬ [𝐴 / 𝑥]𝐵 = 𝐶))
7		df-ne 2379	. . . 4 ⊢ (⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶)
8	7	a1i 9	. . 3 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
9	6, 8	bibi12d 235	. 2 ⊢ (𝐴 ∈ 𝑉 → (([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶) ↔ (¬ [𝐴 / 𝑥]𝐵 = 𝐶 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶)))
10	2, 9	mpbird 167	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ≠ ⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 105   = wceq 1373   ∈ wcel 2178   ≠ wne 2378  [wsbc 3005  ⦋csb 3101

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-fal 1379  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-ne 2379  df-v 2778  df-sbc 3006  df-csb 3102

This theorem is referenced by: (None)