Theorem 4p1e5 9450

Description: 4 + 1 = 5. (Contributed by Mario Carneiro, 18-Apr-2015.)

Assertion

Ref	Expression
4p1e5	|- ( 4 + 1 ) = 5

Proof of Theorem 4p1e5

Step	Hyp	Ref	Expression
1		df-5 9375	. 2 |- 5 = ( 4 + 1 )
2	1	eqcomi 2242	1 |- ( 4 + 1 ) = 5

This proof depends on syntax axioms:   = wceq 1402  (class class class)co 6088   1c1 8186   + caddc 8188   4c4 9366   5c5 9367

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-5 9375

This theorem is used by:  8t7e56  9912  9t6e54  9918  2exp16  13246  prmlem2  13263  163prm  13268  317prm  13269  631prm  13270  1259lem1  13271  1259lem2  13272  1259lem3  13273  1259lem4  13274  log2ublem3  16223  log2ublog2  16224  ex-exp  16946  ex-fac  16947