³x.Vy.Vz.F(x, y, z) = \x.]y.Vz.¬F(x, y, z)

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

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Answer 17

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Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

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Answer 18

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Answer 19

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

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Answer 20

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

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Answer 21

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Answer 22

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Answer 23

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Answer 24

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Answer 25

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Answer 26

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Answer 27

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Answer 28

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Answer 29

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Answer 30

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Answer 31

Follow-up Questions

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Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

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Answer 32

Read through expert solutions to related follow-up questions below.

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

by Bartleby Expert

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Answer 33

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Answer 34

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

by Bartleby Expert

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Answer 35

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Solution

by Bartleby Expert

SEE SOLUTION

Answer 36

Follow-up Question

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Answer 37

Follow-up Question

Answer 38

Follow-up Question

Answer 39

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.

Answer 40

I'm not quite understanding how the right-hand side is not holding up. In the example given, if we fix x at any value, independent of the value y takes, wouldn't there be an equivalence for one value of z? For example, if x=1, then if we choose any value of y we can say 1+y = z. For z = y-1 (and only this value), this equality would be valid and therefore the predicate, ∃x.∀y.∀z x + y ≠ z would be false as there is a value of z that completes the equality going against the for all z quantifier.