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issacnewton

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- Homework Statement
- Prove ##(a+b) + c = a + (b+c)## using Peano postulates

- Relevant Equations
- Peano postulates

I have to prove the associative law for addition ##(a+b) + c = a + (b+c)## using Peano postulates, given that ##a, b, c \in \mathbb{N}##. Now define the set

$$ G = \{ z \in \mathbb{N} |\forall\; x, y \in \mathbb{N} \quad (x + y) + z = x + (y + z) \} $$

Obviously, ## G \subseteq \mathbb{N} ##. Now ## 1 \in \mathbb{N} ## according to Peano postulates. Let ##x, y \in \mathbb{N} ## be arbitrary.

Using the way addition function is defined using the successor function, we have ## (x + y) + 1 = s(x + y) ##. But ##s(x + y) = x + s(y) ## and ## s(y) = y + 1##. So, we have ##(x + y) + 1 = x + s(y) = x + (y + 1)##. This means that ## 1 \in G##.

Now, suppose ##r \in G##. This means that ##r \in \mathbb{N} ## and

$$ \forall\; x, y \in \mathbb{N} \quad (x + y) + r = x + (y + r) $$

Suppose ##x, y \in \mathbb{N} ## be arbitrary. Since, ##r \in G##, we have

$$ (x + y) + r = x + (y + r) $$

$$ \therefore s((x + y) + r) = s( x + (y + r) ) $$

Using definition of addition function, we have

$$ \therefore (x + y) + s(r) = x + s(y + r) = x + (y + s(r)) $$

$$ \therefore (x + y) + s(r) = x + (y + s(r)) $$

From definition of successor function, ## s(r) \in \mathbb{N} ##. Hence ##s(r) \in G##. So, using Peano postulates, ## G = \mathbb{N}##.

Since ##a, b, c \in \mathbb{N}##, we have ## c \in G##. It follows that ##(a+b) + c = a + (b+c)##. Is this proof correct ?

$$ G = \{ z \in \mathbb{N} |\forall\; x, y \in \mathbb{N} \quad (x + y) + z = x + (y + z) \} $$

Obviously, ## G \subseteq \mathbb{N} ##. Now ## 1 \in \mathbb{N} ## according to Peano postulates. Let ##x, y \in \mathbb{N} ## be arbitrary.

Using the way addition function is defined using the successor function, we have ## (x + y) + 1 = s(x + y) ##. But ##s(x + y) = x + s(y) ## and ## s(y) = y + 1##. So, we have ##(x + y) + 1 = x + s(y) = x + (y + 1)##. This means that ## 1 \in G##.

Now, suppose ##r \in G##. This means that ##r \in \mathbb{N} ## and

$$ \forall\; x, y \in \mathbb{N} \quad (x + y) + r = x + (y + r) $$

Suppose ##x, y \in \mathbb{N} ## be arbitrary. Since, ##r \in G##, we have

$$ (x + y) + r = x + (y + r) $$

$$ \therefore s((x + y) + r) = s( x + (y + r) ) $$

Using definition of addition function, we have

$$ \therefore (x + y) + s(r) = x + s(y + r) = x + (y + s(r)) $$

$$ \therefore (x + y) + s(r) = x + (y + s(r)) $$

From definition of successor function, ## s(r) \in \mathbb{N} ##. Hence ##s(r) \in G##. So, using Peano postulates, ## G = \mathbb{N}##.

Since ##a, b, c \in \mathbb{N}##, we have ## c \in G##. It follows that ##(a+b) + c = a + (b+c)##. Is this proof correct ?