Given $\text{a}$ and $\text{b = a-b; a}$ and $\text{b}$ but $\text{c=a+c-b; a}$ or $\text{b=b-a; a}$ but not $\text{b= a+b}$; find $1$ or $(2$ but not $(3$ or $(4$ and $5$ but $(6$ but not $(7$ and $(8$ or $9) ) ) ) ) ).$

- $9$
- $-8$
- $-11$
- $17$

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Given $\text{a}$ and $\text{b = a-b; a}$ and $\text{b}$ but $\text{c=a+c-b; a}$ or $\text{b=b-a; a}$ but not $\text{b= a+b}$; find $1$ or $(2$ but not $(3$ or $(4$ and $5$ but $(6$ but not $(7$ and $(8$ or $9) ) ) ) ) ).$

- $9$
- $-8$
- $-11$
- $17$

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Given that:

- a and b= a-b
- a and b but c= a+c-b
- a or b=b-a
- a but not b=a+b

Now the value of:

1 or (2 but not (3 or (4 and 5 but (6 but not (7 and (8 or 9))))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but (6 but not (7 and (9-8))))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but (6 but not (7 and 1)))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but (6 but not (7 -1)))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but (6 but not 6))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but (6+ 6))))

$\implies$ 1 or (2 but not (3 or (4 and 5 but 12)))

$\implies$ 1 or (2 but not (3 or (4+12-5)))

$\implies$ 1 or (2 but not (3 or 11))

$\implies$ 1 or (2 but not (11-3))

$\implies$ 1 or (2 but not 8)

$\implies$ 1 or (8+2)

$\implies$ 1 or 10

$\implies 10-1=9$

So option (A) is correct.

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