Theorem necom 2598

Description: Commutation of inequality. (Contributed by NM, 14-May-1999.)

Assertion

Ref	Expression
necom	⊢ (A ≠ B ↔ B ≠ A)

Proof of Theorem necom

Step	Hyp	Ref	Expression
1		eqcom 2355	. 2 ⊢ (A = B ↔ B = A)
2	1	necon3bii 2549	1 ⊢ (A ≠ B ↔ B ≠ A)

This proof depends on syntax axioms:   ↔ wb 176   ≠ wne 2517

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-cleq 2346  df-ne 2519

This theorem is used by:  necomi  2599  necomd  2600  0pss  3589  difprsn1  3848  difprsn2  3849  diftpsn3  3850  fvpr1  5450  fvpr2  5451