Theorem mteqand 3020

Description: A modus tollens deduction for inequality. (Contributed by Steven Nguyen, 1-Jun-2023.)

Hypotheses

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

Assertion

Ref	Expression
mteqand	⊢ (𝜑 → 𝐴 ≠ 𝐵)

Proof of Theorem mteqand

Step	Hyp	Ref	Expression
1		mteqand.1	. . . 4 ⊢ (𝜑 → 𝐶 ≠ 𝐷)
2	1	neneqd 2934	. . 3 ⊢ (𝜑 → ¬ 𝐶 = 𝐷)
3		mteqand.2	. . 3 ⊢ ((𝜑 ∧ 𝐴 = 𝐵) → 𝐶 = 𝐷)
4	2, 3	mtand 815	. 2 ⊢ (𝜑 → ¬ 𝐴 = 𝐵)
5	4	neqned 2936	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1541   ≠ wne 2929

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-an 396  df-ne 2930

This theorem is referenced by:  isdrngd  20682  imadrhmcl  20714  fracfld  33281  qsidomlem2  33425  rprmasso  33497  vr1nz  33561  rtelextdg2lem  33760  2sqr3minply  33814  cos9thpiminplylem2  33817  zarcmplem  33915  expeq1d  42443  remul01  42526  remulinvcom  42552  mulgt0b2d  42597  sn-inelr  42606  ricdrng1  42647  prjspersym  42726  prjspreln0  42728  prjspner1  42745  flt0  42756  fltne  42763  eufunc  49648