Theorem necon3bii 2405

Description: Inference from equality to inequality. (Contributed by NM, 23-Feb-2005.)

Hypothesis

Ref	Expression
necon3bii.1	⊢ (𝐴 = 𝐵 ↔ 𝐶 = 𝐷)

Assertion

Ref	Expression
necon3bii	⊢ (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷)

Proof of Theorem necon3bii

Step	Hyp	Ref	Expression
1		necon3bii.1	. . 3 ⊢ (𝐴 = 𝐵 ↔ 𝐶 = 𝐷)
2	1	necon3abii 2403	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐶 = 𝐷)
3		df-ne 2368	. 2 ⊢ (𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷)
4	2, 3	bitr4i 187	1 ⊢ (𝐴 ≠ 𝐵 ↔ 𝐶 ≠ 𝐷)

Syntax hints:  ¬ wn 3   ↔ wb 105   = wceq 1364   ≠ wne 2367

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

This theorem depends on definitions:  df-bi 117  df-ne 2368

This theorem is referenced by:  necom  2451  negne0bi  8299  3dvds  12029