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Question 6.5: Is the statement ∃x ∀y [x+y=y] a member of Th(N,+)? Why or w...

Is the statement ∃x ∀y [x+y=y] a member of Th(N,+)? Why or why not? What about the statement ∃x ∀y [x+y=x]?

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The statement ∃x ∀y [x+y=y] is a member of Th(N,+) because that statement is true for the standard interpretation of + over the universe N. Recall that we use N = {0, 1, 2, . . .} in this chapter and so we may use x =0.

The statement ∃x ∀y [x+y=x] is not a member of Th(N,+) because that statement isn’t true in this model. For any value of x, setting y = 1 causes x+y=x to fail.

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