Theorem sbcne12g 3110

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 3108	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
2	1	notbid 668	. 2 ⊢ (𝐴 ∈ 𝑉 → (¬ [𝐴 / 𝑥]𝐵 = 𝐶 ↔ ¬ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
3		df-ne 2376	. . . . 5 ⊢ (𝐵 ≠ 𝐶 ↔ ¬ 𝐵 = 𝐶)
4	3	sbcbii 3057	. . . 4 ⊢ ([𝐴 / 𝑥]𝐵 ≠ 𝐶 ↔ [𝐴 / 𝑥] ¬ 𝐵 = 𝐶)
5		sbcng 3038	. . . 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 2376	. . . 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 1372   ∈ wcel 2175   ≠ wne 2375  [wsbc 2997  ⦋csb 3092

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 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-fal 1378  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-ne 2376  df-v 2773  df-sbc 2998  df-csb 3093

This theorem is referenced by: (None)