Theorem 4

$$\textrm{Tr}[\gamma_5] = \textrm{Tr}[i\gamma^0\gamma^1\gamma^2\gamma^3]$$

$$= -\textrm{Tr}[i\gamma^1\gamma^0\gamma^2\gamma^3]$$

$$= \textrm{Tr}[i\gamma^1\gamma^2\gamma^0\gamma^3]$$

$$= -\textrm{Tr}[i\gamma^1\gamma^2\gamma^3\gamma^0]$$

$$= -\textrm{Tr}[\gamma_5]$$

$$= 0 ,$$ (7.65)

This leads to

$$\textrm{Tr}[\gamma_5\not{a}\not{b}] = 0 ,$$ (7.66)

$$\textrm{Tr}[\gamma_5\not{a}\not{b}\not{c}\not{d}] = 4 i \epsilon_{\alpha\beta\gamma\delta} a^\alpha b^\beta c^\gamma d^\delta ,$$ (7.67)

where $\epsilon_{\alpha\beta\gamma\delta}$ is $+1$ for $(\alpha,\beta,\gamma,\delta)$, an even permutation of $(0,1,2,3)$ and is $-1$ for an odd permutation, and is 0 it two indices are the same

Proof: For a non-vanishing contribution all components of $a,b,c,d$ must be different and the total contribution is the sum of the various combinations of components multiplied by the sign of the permutation. To fix the overall sign take an example case:

$$\textrm{Tr}[\gamma_5\gamma_0\gamma_1\gamma_2\gamma_3a^0b^1c^2d^3] = i\epsilon_{0123} a^0b^1c^2d^3 \textrm{Tr}[\gamma_5^2]$$

$$= 4i\epsilon_{0123} a^0b^1c^2d^3.$$ (7.68)