Theorem mteqand 3122

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 3021	. . 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 3023	1 ⊢ (𝜑 → 𝐴 ≠ 𝐵)

Syntax hints:   → wi 4   ∧ wa 398   = wceq 1537   ≠ wne 3016

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 209  df-an 399  df-ne 3017

This theorem is referenced by:  qsidomlem2  30966  uvcn0  39200  remul01  39286  remulinvcom  39297  prjspersym  39306  prjspreln0  39308  fltne  39321