Theorem necom 1198

Description: Commutation of inequality.

Assertion

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

Proof of Theorem necom

Step	Hyp	Ref	Expression
1		cleqcom 1103	. . 3 ⊢ (A = B ↔ B = A)
2	1	negbii 162	. 2 ⊢ (¬ A = B ↔ ¬ B = A)
3		df-ne 1192	. 2 ⊢ (A ≠ B ↔ ¬ A = B)
4		df-ne 1192	. 2 ⊢ (B ≠ A ↔ ¬ B = A)
5	2, 3, 4	3bitr4 158	1 ⊢ (A ≠ B ↔ B ≠ A)

Syntax hints:  ¬ wn 1   ↔ wb 127   = wceq 1091   ≠ wne 1190

This theorem is referenced by:  gt0ne0 4340  recgt0i 4385

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-cleq 1097  df-ne 1192

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