Theorem 5p1e6 9451

Description: 5 + 1 = 6. (Contributed by Mario Carneiro, 18-Apr-2015.)

Assertion

Ref	Expression
5p1e6	|- ( 5 + 1 ) = 6

Proof of Theorem 5p1e6

Step	Hyp	Ref	Expression
1		df-6 9376	. 2 |- 6 = ( 5 + 1 )
2	1	eqcomi 2242	1 |- ( 5 + 1 ) = 6

This proof depends on syntax axioms:   = wceq 1402  (class class class)co 6088   1c1 8186   + caddc 8188   5c5 9367   6c6 9368

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-6 9376

This theorem is used by:  8t8e64  9913  9t7e63  9919  5recm6rec  9936  fldiv4p1lem1div2  10761  5ndvds6  12727  163prm  13268  631prm  13270  1259lem1  13271  1259lem4  13274  log2ublem3  16223  log2ublog2  16224