Theorem mteqand 3024

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 2938	. . 3 ⊢ (𝜑 → ¬ 𝐶 = 𝐷)
3		mteqand.2	. . 3 ⊢ ((𝜑 ∧ 𝐴 = 𝐵) → 𝐶 = 𝐷)
4	2, 3	mtand 816	. 2 ⊢ (𝜑 → ¬ 𝐴 = 𝐵)
5	4	neqned 2940	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1542   ≠ wne 2933

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 2934

This theorem is referenced by:  isdrngd  20733  imadrhmcl  20765  fracfld  33384  qsidomlem2  33528  rprmasso  33600  vr1nz  33668  rtelextdg2lem  33886  2sqr3minply  33940  cos9thpiminplylem2  33943  zarcmplem  34041  expeq1d  42770  remul01  42853  remulinvcom  42879  mulgt0b2d  42937  sn-inelr  42946  ricdrng1  42987  prjspersym  43054  prjspreln0  43056  prjspner1  43073  flt0  43084  fltne  43091  eufunc  50009