NOR and NAND operator tautology

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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?

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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 $\downarrow$.

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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.

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There seem to be many parts to what you're asking so I'll just break down your question:

"$A↓(A↓A) \equiv T$"

is wrong.

Instead it should be

$A↓(A↓A) \equiv F$

which is an always-false statement. This is the opposite of a tautology and is known as a contradiction.

"we know that $(A↓A) \equiv \sim A$ where $ ↓$ is NOR symbol."

Correct.

"so $A↓(A↓A) \equiv A↓ \sim A \equiv \sim(A \lor \sim A) \equiv F$"

All correct.

"but the book said that since we know $(A↓A) \equiv \sim A \equiv (A|A) $ where | is NAND symbol"

This is correct and holds when both operands are identical.

"$A↓(A↓A) \equiv A| \sim A...$"

Wrong. It is unclear how you made that leap.

"$...\equiv \sim (A \land \sim A) \equiv\sim F \equiv T$"

The rest of this chain of reasoning (after the wrong step) is correct but obviously the conclusion is wrong because of that earlier mistake.

I hope that clarifies things.