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Yoav Kallus
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Trying to figure out the logic in which the following formula is expressible: \forall i\in N: (x_i > y_i)$\forall i\in N: (x_i > y_i)$, which is equivalent to the "infinite" conjunction \wedge_{i\in N} (x_i > y_i)$\bigwedge_{i\in N} (x_i > y_i)$.

Now a 1st order logic allows arbitrary number of variables {x_i,y_i|i\in N}$\{x_i,y_i\mid i\in N\}$, but only a finite number of atomic formulas can be composed. (Here x_i > y_i$x_i > y_i$ is an atomic formula based on binary predicate ">""$>$", and so the above formula is a composition of infinite number of atomic formulas.) Thus in the form written, the above doesn't seem to be a formula of 1st-order logic. Also, it's not clear how to rewrite this formula in 1st-order logic if indeed it belongs there.

Next in 2nd-order logic, that allows quantification over predicates (as well as functions), again it is not clear whether the above can be written as formula in 2nd-order logic.

Any insights? Thanks.

Trying to figure out the logic in which the following formula is expressible: \forall i\in N: (x_i > y_i), which is equivalent to the "infinite" conjunction \wedge_{i\in N} (x_i > y_i).

Now a 1st order logic allows arbitrary number of variables {x_i,y_i|i\in N}, but only a finite number of atomic formulas can be composed. (Here x_i > y_i is an atomic formula based on binary predicate ">", and so the above formula is a composition of infinite number of atomic formulas.) Thus in the form written, the above doesn't seem to be a formula of 1st-order logic. Also, it's not clear how to rewrite this formula in 1st-order logic if indeed it belongs there.

Next in 2nd-order logic, that allows quantification over predicates (as well as functions), again it is not clear whether the above can be written as formula in 2nd-order logic.

Any insights? Thanks.

Trying to figure out the logic in which the following formula is expressible: $\forall i\in N: (x_i > y_i)$, which is equivalent to the "infinite" conjunction $\bigwedge_{i\in N} (x_i > y_i)$.

Now a 1st order logic allows arbitrary number of variables $\{x_i,y_i\mid i\in N\}$, but only a finite number of atomic formulas can be composed. (Here $x_i > y_i$ is an atomic formula based on binary predicate "$>$", and so the above formula is a composition of infinite number of atomic formulas.) Thus in the form written, the above doesn't seem to be a formula of 1st-order logic. Also, it's not clear how to rewrite this formula in 1st-order logic if indeed it belongs there.

Next in 2nd-order logic, that allows quantification over predicates (as well as functions), again it is not clear whether the above can be written as formula in 2nd-order logic.

Any insights? Thanks.

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What logic can express this sentence?

Trying to figure out the logic in which the following formula is expressible: \forall i\in N: (x_i > y_i), which is equivalent to the "infinite" conjunction \wedge_{i\in N} (x_i > y_i).

Now a 1st order logic allows arbitrary number of variables {x_i,y_i|i\in N}, but only a finite number of atomic formulas can be composed. (Here x_i > y_i is an atomic formula based on binary predicate ">", and so the above formula is a composition of infinite number of atomic formulas.) Thus in the form written, the above doesn't seem to be a formula of 1st-order logic. Also, it's not clear how to rewrite this formula in 1st-order logic if indeed it belongs there.

Next in 2nd-order logic, that allows quantification over predicates (as well as functions), again it is not clear whether the above can be written as formula in 2nd-order logic.

Any insights? Thanks.