Theorem mteqand 3026

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 2940	. . 3 ⊢ (𝜑 → ¬ 𝐶 = 𝐷)
3		mteqand.2	. . 3 ⊢ ((𝜑 ∧ 𝐴 = 𝐵) → 𝐶 = 𝐷)
4	2, 3	mtand 821	. 2 ⊢ (𝜑 → ¬ 𝐴 = 𝐵)
5	4	neqned 2942	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 396   = wceq 1547   ≠ wne 2935

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 208  df-an 397  df-ne 2936

This theorem is referenced by:  isdrngd  20744  imadrhmcl  20776  fracfld  33399  qsidomlem2  33543  rprmasso  33615  vr1nz  33683  rtelextdg2lem  33917  2sqr3minply  33971  cos9thpiminplylem2  33974  zarcmplem  34072  expeq1d  42808  remul01  42891  remulinvcom  42917  mulgt0b2d  42975  sn-inelr  42984  ricdrng1  43021  prjspersym  43064  prjspreln0  43066  prjspner1  43083  flt0  43094  fltne  43101  eufunc  50019