Immediate consequence of a theorem related to spanning set/basis - linear algebra

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I am going through a theorem and its corollary from a text book:

Theorem $T_0$: Let $W$ be a subspace of $R^n$, and let $B = \{v_1,...,v_p\}$ be a spanning set of $W$ containing $p$ vectors. Then any set of $p+1$ or more vectors in $W$ is a linearly dependent set.

Corollary to the above theorem : Let $W$ be subspace of $R^n$, and let $B = \{w_1,...,w_p\}$ be a basis of $W$ containing $p$ vectors. Then every basis for $W$ contains $p$ vectors.

Theorem to be proved $T_1$: Any set of fewer than $p$ vectors in $W$ does not span $W$.

The textbook claims that : $T_1$ is equivalent to the statement that a spanning set for $W$ must contain at least $p$ vectors. Again, this is an immediate consequence of theorem $T_0$.

I could not understand why it is an immediate consequence. Can somebody help me out.

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Remember that a basis of a vector space is a linearly independent set and what generates that vector space. Furthermore, All bases for a finite-dimensional vector space have the same number of vectors.