Theorem mteqand 3023

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 2937	. . 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 2939	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1541   ≠ wne 2932

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 2933

This theorem is referenced by:  isdrngd  20698  imadrhmcl  20730  fracfld  33390  qsidomlem2  33534  rprmasso  33606  vr1nz  33674  rtelextdg2lem  33883  2sqr3minply  33937  cos9thpiminplylem2  33940  zarcmplem  34038  expeq1d  42575  remul01  42658  remulinvcom  42684  mulgt0b2d  42729  sn-inelr  42738  ricdrng1  42779  prjspersym  42846  prjspreln0  42848  prjspner1  42865  flt0  42876  fltne  42883  eufunc  49763