I am having trouble with a problem in the book I studied about logic that use the NOR operator (also known as Peirce's arrow) and the NAND operator.
The discrete math book said this is tautology $A↓(A↓A) \equiv T$.
We know that $(A↓A) \equiv \mathord\sim A$ where $ ↓$ is the NOR symbol.
so $A↓(A↓A) \equiv A↓ \mathord\sim A \equiv \mathord\sim(A \lor \mathord\sim A) \equiv F$
but the book said that since we know $(A↓A) \equiv \mathord\sim A \equiv (A|A) $ where | is NAND symbol
$A↓(A↓A) \equiv A| \mathord\sim A \equiv \mathord\sim (A \land \mathord\sim A) \equiv\mathord\sim F \equiv T$ so it is a tautology.
So is it a tautology or not?
Best Answer
A statement is a tautology if it is true under every possible interpretations of the literals that form it. Here is the truth table for $\downarrow$.
So if $A$ is true, then $$A\downarrow(A\downarrow A) \equiv T\downarrow(T\downarrow T)\equiv T\downarrow F\equiv F$$ and if $A$ is false then $$A\downarrow(A\downarrow A) \equiv F\downarrow(F\downarrow F)\equiv F\downarrow T\equiv F$$
So, not only is $A\downarrow(A\downarrow A)$ not a tautology, it is a contradiction.