Theorem mteqand 3019

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 2933	. . 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 2935	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1541   ≠ wne 2928

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 2929

This theorem is referenced by:  isdrngd  20678  imadrhmcl  20710  fracfld  33269  qsidomlem2  33413  rprmasso  33485  vr1nz  33549  rtelextdg2lem  33734  2sqr3minply  33788  cos9thpiminplylem2  33791  zarcmplem  33889  expeq1d  42356  remul01  42439  remulinvcom  42465  mulgt0b2d  42510  sn-inelr  42519  ricdrng1  42560  prjspersym  42639  prjspreln0  42641  prjspner1  42658  flt0  42669  fltne  42676  eufunc  49553