Theorem mteqand 3016

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 2930	. . 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 2932	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1540   ≠ wne 2925

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 2926

This theorem is referenced by:  isdrngd  20674  imadrhmcl  20706  fracfld  33258  qsidomlem2  33424  rprmasso  33496  vr1nz  33559  rtelextdg2lem  33716  2sqr3minply  33770  cos9thpiminplylem2  33773  zarcmplem  33871  expeq1d  42312  remul01  42395  remulinvcom  42421  mulgt0b2d  42466  sn-inelr  42475  ricdrng1  42516  prjspersym  42595  prjspreln0  42597  prjspner1  42614  flt0  42625  fltne  42632  eufunc  49511