Autosave: 2022-12-18 14:30:04

Logic/Laws_and_theorems.md/DeMorgan's_Laws.md

@@ -1,9 +1,11 @@
 ---
 categories:
-  - Mathematics
-tags: [logic, theorems]
+  - Logic
+tags: [logic, laws]
 ---
 
+# DeMorgan's Laws
+
 DeMorgan's laws express some fundamental equivalences that obtain between the Boolean [connectives](Truth-functional%20connectives.md):
 
 ## First Law
@@ -16,14 +18,14 @@ $$
 
 The equivalence is demonstrated with the following truth-table
 
-![demorgan-1.png](../img/demorgan-1.png)
+![demorgan-1.png](/img/demorgan-1.png)
 
 ## Second Law
 
 > The negation of a disjunction is equivalent to the conjunction of the negation of the original disjuncts.
 
 $$
-\sim (P \lor Q) \equiv \sim P & \sim Q
+\sim (P \lor Q) \equiv \sim P \& \sim Q
 $$
 
-![demorgan-2.png](../img/demorgan-2.png)
+![demorgan-2.png](/img/demorgan-2.png)

Logic/Propositional_logic/Boolean_algebra.md

@@ -6,9 +6,11 @@ tags: [propositional-logic, algebra]
 
 # Boolean algebra
 
+## Algebraic laws
+
 Many of the laws that obtain in the mathematical realm of algebra also obtain for Boolean expressions.
 
-## The Commutative Law
+### The Commutative Law
 
 $$
     x \land y = y \land x \\
@@ -21,7 +23,7 @@ $$
 
 Compare the [Commutative Law](/Mathematics/Prealgebra/Whole_numbers.md#the-commutative-property) in the context of arithmetic.
 
-## The Associative Law
+### The Associative Law
 
 $$
     x   \land (y \land z) = (x \land y) \land z
@@ -33,7 +35,7 @@ $$
 
 Compare the [Associative Law](/Mathematics/Prealgebra/Whole_numbers.md#the-associative-property) in the context of arithmetic.
 
-## The Distributive Law
+### The Distributive Law
 
 $$
     x \land (y \lor z) = (x \land y) \lor (x \land z)
@@ -50,3 +52,54 @@ $$
 $$
 
 In addition we have [DeMorgan's Laws](/Logic/Laws_and_theorems.md/DeMorgan's_Laws.md) which express the relationship that obtains between the negations of conjunctive and disjunctive expressions
+
+## Applying the laws to simplify complex Boolean expressions
+
+Say we have the following expression:
+
+$$
+    \lnot(\lnot(x) \land \lnot (x \lor y))
+$$
+
+We can employ DeMorgan's Laws to convert the second conjunct to a different form:
+
+$$
+    \lnot (x \lor x) = \lnot x \land \lnot y
+$$
+
+So now we have:
+
+$$
+    \lnot(\lnot(x) \land (\lnot x \land \lnot y ))
+$$
+
+As we have now have an expression of the form _P and (Q and R)_ we can apply the Distributive Law to simplify the brackets (_P and Q and R_):
+
+$$
+    \lnot( \lnot(x) \land \lnot(x) \land \lnot(y))
+$$
+
+Notice that we are repeating ourselves in this reformulation. We have $\lnot(x) \land \lnot(x)$ but this is just the same $\lnot(x)$ by the principle of **idempotence**. So we can reduce to:
+
+$$
+    \lnot(\lnot(x) \land \lnot(y))
+$$
+
+This gives our expression the form of the first DeMorgan Law ($\lnot (P \land Q)$), thus we can apply the law ($\lnot P \lor \lnot Q$) to get:
+
+$$
+\lnot(\lnot(x)) \lor \lnot(\lnot(y))
+$$
+
+Of course now we have two double negatives. We can apply the double negation law to:
+
+$$
+    x \lor y
+$$
+
+// TO DO:
+
+- Use truth tables to show equivalence
+- Explicitly add implicit laws
+- Link to deductive rules
+- Link to digital circuits and NANDs as universal gates