Theorem mteqand 3022

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 2934	. . 3 ⊢ (𝜑 → ¬ 𝐶 = 𝐷)
3		mteqand.2	. . 3 ⊢ ((𝜑 ∧ 𝐴 = 𝐵) → 𝐶 = 𝐷)
4	2, 3	mtand 814	. 2 ⊢ (𝜑 → ¬ 𝐴 = 𝐵)
5	4	neqned 2936	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 394   = wceq 1533   ≠ wne 2929

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 206  df-an 395  df-ne 2930

This theorem is referenced by:  isdrngd  20697  imadrhmcl  20725  fracfld  33146  qsidomlem2  33316  rprmasso  33388  2sqr3minply  33575  zarcmplem  33664  ricdrng1  41960  remul01  42141  remulinvcom  42166  prjspersym  42210  prjspreln0  42212  prjspner1  42229  flt0  42240  fltne  42247