Theorem necomd 2600

Description: Deduction from commutative law for inequality. (Contributed by NM, 12-Feb-2008.)

Hypothesis

Ref	Expression
necomd.1	|- (ph -> A =/= B)

Assertion

Ref	Expression
necomd	|- (ph -> B =/= A)

Proof of Theorem necomd

Step	Hyp	Ref	Expression
1		necomd.1	. 2 |- (ph -> A =/= B)
2		necom 2598	. 2 |- (A =/= B <-> B =/= A)
3	1, 2	sylib 188	1 |- (ph -> B =/= A)

This proof depends on syntax axioms:   -> wi 4   =/= 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:  difsnb  3851  vfinncvntnn  4549  vfinncvntsp  4550  nchoicelem12  6301  nchoicelem17  6306