Search
Search Icon Icon to open search Reverse Ordering Last updated
Nov 1, 2022
Table of Contents Proof Proof # Statement 1Let ( P , ≤ ) (P, \leq) ( P , ≤ ) be a Partial Ordering . Then the Reverse Ordering of P P P , ( P , ≤ ′ ) (P, \leq’) ( P , ≤ ′ ) is defined as a ≤ ′ b ⇔ a ≥ b a \leq’ b \Leftrightarrow a \geq b a ≤ ′ b ⇔ a ≥ b ∀ a , b ∈ T \forall a,b \in T ∀ a , b ∈ T . It is a Partial Ordering .
# ProofFor all a ∈ P a \in P a ∈ P , a ≤ ′ a a \leq’ a a ≤ ′ a since a ≥ a a \geq a a ≥ a ✓ \checkmark ✓ . Suppose a , b ∈ P a,b \in P a , b ∈ P . If a ≤ ′ b a \leq’ b a ≤ ′ b and b ≤ ′ a b \leq’ a b ≤ ′ a , then a ≥ b a \geq b a ≥ b and b ≥ a b \geq a b ≥ a , so a = b a = b a = b ✓ \checkmark ✓ . Suppose a , b , c ∈ P a,b,c \in P a , b , c ∈ P . If a ≤ ′ b a \leq’ b a ≤ ′ b and b ≤ ′ c b \leq’ c b ≤ ′ c , then a ≥ b a \geq b a ≥ b and b ≥ c b \geq c b ≥ c , so a ≥ c a \geq c a ≥ c , implying that a ≤ ′ c a \leq’ c a ≤ ′ c ✓ \checkmark ✓ . ■ \blacksquare ■
# Statement 2The Reverse Ordering of a Total Ordering is a Total Ordering .
# ProofThis follows because we can compare any a , b ∈ T a,b \in T a , b ∈ T with ≤ ′ \leq’ ≤ ′ using the corresponding comparison with ≤ \leq ≤ . ■ \blacksquare ■