R, 86 84 bytes

-1 byte by using !m instead of m<1; -1 byte by realizing I only need one long OR || (the first one can be a short OR |).

a=function(x,m=max(x),B=F)!m|all(x-1)==B||any(sapply(1:m,function(y)!a(x-y,y-1,!B)))

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Exhaustive recursive search. Takes input as a vector. Outputs TRUE or FALSE.

At each step, subtracts the value chosen by the player from all the values in the set x.

Alice wins if at her turn, the updated x includes 1. Bob wins if at his turn, the updated x does not include 1. To this end, the function includes a boolean B, worth FALSE when Alice plays and TRUE when Bob plays, and checks all(x != 1) == B. If this condition is verified, then the player wins, else the player wins if their is a move which makes the other player lose (last condition in the code).

The initial !m (equivalent to m!=0) is there to check whether at the last move, the other player played 1 even though it is a losing move.

Edit: Arnauld added an explanation to his JavaScript answer while I was typing; it seems we have similar algorithms.

R, 86 83 bytes

-1 byte by using !m instead of m<1; -1 byte by realizing I only need one long OR || (the first one can be a short OR |); -1 byte by toggling B and using subtraction instead of equality of booleans.

a=function(x,m=max(x),B=T)!m|all(x-1)-B||any(sapply(1:m,function(y)!a(x-y,y-1,!B)))

Try it online!

Exhaustive recursive search. Takes input as a vector. Outputs TRUE or FALSE.

At each step, subtracts the value chosen by the player from all the values in the set x.

Alice wins if at her turn, the updated x includes 1. Bob wins if at his turn, the updated x does not include 1. To this end, the function includes a boolean B, worth TRUE when Alice plays and FALSE when Bob plays, and checks all(x != 1) != B. If this condition is verified then the player wins, else the player wins if there is a move which makes the other player lose (last condition in the code, uses recursion).

The initial !m (equivalent to m!=0) is there to check whether at the last move, the other player played 1 even though it is a losing move.

Edit: Arnauld added an explanation to his JavaScript answer while I was typing; it seems we have similar algorithms.

R, 86 bytes

a=function(x,m=max(x),B=F)m<1||all(x-1)==B||any(sapply(1:m,function(y)!a(x-y,y-1,!B)))

Try it online!

Exhaustive recursive search. Takes input as a vector. Outputs TRUE or FALSE.

At each step, subtracts the value chosen by the player from all the values in the set x.

Alice wins if at her turn, the updated x includes 1. Bob wins if at his turn, the updated x does not include 1. To this end, the function includes a boolean B, worth FALSE when Alice plays and TRUE when Bob plays, and checks all(x != 1) == B. If this condition is verified, then the player wins, else the player wins if their is a move which makes the other player lose (last condition in the code).

The initial m<1 is there to check whether at the last move, the other player played 1 even though it is a losing move.

Edit: Arnauld added an explanation to his JavaScript answer while I was typing; it seems we have the same algorithm.

R, 86 84 bytes

-1 byte by using !m instead of m<1; -1 byte by realizing I only need one long OR || (the first one can be a short OR |).

a=function(x,m=max(x),B=F)!m|all(x-1)==B||any(sapply(1:m,function(y)!a(x-y,y-1,!B)))

Try it online!

Exhaustive recursive search. Takes input as a vector. Outputs TRUE or FALSE.

At each step, subtracts the value chosen by the player from all the values in the set x.

Alice wins if at her turn, the updated x includes 1. Bob wins if at his turn, the updated x does not include 1. To this end, the function includes a boolean B, worth FALSE when Alice plays and TRUE when Bob plays, and checks all(x != 1) == B. If this condition is verified, then the player wins, else the player wins if their is a move which makes the other player lose (last condition in the code).

The initial !m (equivalent to m!=0) is there to check whether at the last move, the other player played 1 even though it is a losing move.

Edit: Arnauld added an explanation to his JavaScript answer while I was typing; it seems we have similar algorithms.