Theorem 3netr4d 2400

Description: Substitution of equality into both sides of an inequality. (Contributed by NM, 24-Jul-2012.)

Hypotheses

Ref	Expression
3netr4d.1	⊢ (𝜑 → 𝐴 ≠ 𝐵)
3netr4d.2	⊢ (𝜑 → 𝐶 = 𝐴)
3netr4d.3	⊢ (𝜑 → 𝐷 = 𝐵)

Assertion

Ref	Expression
3netr4d	⊢ (𝜑 → 𝐶 ≠ 𝐷)

Proof of Theorem 3netr4d

Step	Hyp	Ref	Expression
1		3netr4d.1	. 2 ⊢ (𝜑 → 𝐴 ≠ 𝐵)
2		3netr4d.2	. . 3 ⊢ (𝜑 → 𝐶 = 𝐴)
3		3netr4d.3	. . 3 ⊢ (𝜑 → 𝐷 = 𝐵)
4	2, 3	neeq12d 2387	. 2 ⊢ (𝜑 → (𝐶 ≠ 𝐷 ↔ 𝐴 ≠ 𝐵))
5	1, 4	mpbird 167	1 ⊢ (𝜑 → 𝐶 ≠ 𝐷)

Syntax hints:   → wi 4   = 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  ax-5 1461  ax-gen 1463  ax-4 1524  ax-17 1540  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-cleq 2189  df-ne 2368

This theorem is referenced by:  modsumfzodifsn  10488  ennnfonelemnn0  12639  lgsne0  15279