Common use of Proof Clause in Contracts

Proof. With f = 1 and n 4 we have at least 3 correct nodes. A correct node will see every correct value at least twice, once directly from another correct node, and once through the third correct node. So all correct values are in T . If the byzantine node sends the same value to at least 2 other (correct) nodes, all correct nodes will see the value twice, so all add it to set T . If the byzantine node sends all different values to the correct nodes, none of these values will end up in any set T .

Proof. With f = 1 and n 4 we have at least 3 correct nodes. A correct node will see every correct value at least twice, once directly from another correct node, and once through the third correct node. So all correct values are in T . If the byzantine node sends the same value to at least 2 other (correct) nodes, all correct nodes will see the value twice, so all add it to set T . If the byzantine node sends all different values to the correct nodes, none of these values will end up in any set T . Theorem 3.11. Algorithm 3.9 reaches byzantine agreement if n ≥ 4.

Proof. With f = 1 and n 4 we have at least 3 correct nodes. A correct node will see every correct value at least twice, once directly from another correct node, and once through the third correct node. So all correct values are in T . If the byzantine node sends the same value to at least 2 other (correct) nodes, all correct nodes will see the value twice, so all add it to set T . If the byzantine node sends all different values to the correct nodes, none of these values will end up in any set T .