Pions, parity, spin

Pions have odd parity ($P=-1$) which means their wavefunction is anti-symmetric $\psi(x)=-\psi(-x)$. According to Spin-Statistics theorem fermions (spin 1/2 particles) have anti-symmetric wavefunctions. Just looking at the wavefunction it seems that pions are fermions. However, we know that pions have spin 0 or spin 1 and thus are bosons.

There must be something wrong in the above chain of conclusion. What is it?

Fermion wavefunctions are antisymmetric under the interchange of two particles. Spatial inversion flips the spatial coordinate, but does not interchange particles.

In other words, let's say we have a two particle wave function, $\psi(x_1, x_2)$ (where $x_1$ is the position of particle 1, and $x_2$ is the position of particle 2).

Being odd under parity says: $$\psi(x_1,x_2) = - \psi(-x_1,-x_2).$$

Being odd under interchange of particles says $$\psi(x_1, x_2) = - \psi(x_2, x_1).$$

Thus parity and statistics are independent properties. In particular, it is perfectly consistent to have a parity odd boson.

(Things get a little more interesting if you have spin, because parity also affects the polarization, but that seems like a more complicated question than what you asked).

The pion is a pseudoscalar particle, which behaves like a scalar, except that it changes sign under a parity inversion while a true scalar does not.

For details, see this post by @Luboš Motl and links there: What is a Pseudoscalar particle?

well, fermions' "spatial wave function" can also be antisymmetric. I think it's the whole wave function(spin+spatial) that matters.