diff --git a/SM3中的FF2和GG2函数.md b/SM3中的FF2和GG2函数.md
index d3502e9..541a9b0 100644
--- a/SM3中的FF2和GG2函数.md
+++ b/SM3中的FF2和GG2函数.md
@@ -39,4 +39,12 @@ GG2等价公式初次见于[Intel® Integrated Performance Primitives Cryptograp
 |1|1|1|1|1|1|1|
 
 # 证明
-Ask help https://math.stackexchange.com/questions/4775054/how-to-prove-below-two-logic-formulas
\ No newline at end of file
+Ask help https://math.stackexchange.com/questions/4775054/how-to-prove-below-two-logic-formulas  
+GG2(X,Y,Z)
+= $(X \land Y) \bigoplus (\lnot X \land Z)$  
+= $(\lnot (X \land Y) \land (\lnot X \land Z)) \lor ((X \land Y) \land (\lnot (\lnot X \land Z)))$   
+= $((\lnot X \lor \lnot Y) \land (\lnot X \land Z)) \lor ((X \land Y) \land (X \lor \lnot Z))$  
+= $(\lnot X \land Z) \lor (\lnot X \land \lnot Y \land Z) \lor (X \land Y) \lor (X \land Y \land \lnot Z) $  
+= $((\lnot X \land Z) \lor (\lnot X \land Z \land \lnot Y)) \lor ((X \land Y) \lor (X \land Y \land \lnot Z))$  
+= $((X \land Y) \land (1 \lor \lnot Z)) \lor ((\lnot X \land Z) \land (1 \lor \lnot Y))$  
+= $(X \land Y) \lor (\lnot X \land Z)$  
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