= Solution
The <Dirac field> has the global <symmetry> $\psi\mapsto e^{i\alpha}\psi$, $\bar\psi\mapsto e^{-i\alpha}\bar\psi$, while the real scalar is unchanged. Both the free <Dirac action> and the <pseudoscalar Yukawa interaction> preserve this <symmetry>. Therefore <Dirac fermion number conservation> holds: its charge counts particles minus <antiparticles>.
The initial state has charge $+1$, whereas a final <antiparticle> and a neutral scalar have charge $-1$. Since the <S-matrix> commutes with that charge,
$$
(Q_f-Q_i)\langle f|S|i\rangle=0
\quad\Longrightarrow\quad
\boxed{T(f\phi\to\bar f\phi)=0,\qquad\sigma=0}.
$$
This holds at every order, not just tree level. In the <Feynman rules>, a continuous fermion arrow cannot connect these specified external states. Replacing only the outgoing $\bar u$ by a $v$ in the previous expression would not give a physical amplitude. Changing both external <fermions> to <antiparticles> would instead produce an allowed process with reversed fermion flow and the appropriate $v$ spinors; moving a leg between initial and final states is a different operation governed by <crossing symmetry>.
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