= Solution
Use $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and $\hbar=c=1$. In the <interaction picture>, fields evolve with the free <Hamiltonian>, while interactions enter the state evolution through the <Dyson series>. Write the <Klein-Gordon field> as $\phi=\phi^{(+)}+\phi^{(-)}$, with the <positive-frequency part of a quantum field> containing <annihilation operators> and $\phi^{(-)}=(\phi^{(+)})^\dagger$ containing <creation operators>. The mode normalization in the PDF corresponds to $[a_{\mathbf p},a^\dagger_{\mathbf q}]=(2\pi)^3\delta^3(\mathbf p-\mathbf q)$ and $E_{\mathbf p}=\sqrt{|\mathbf p|^2+m^2}$.
The <time-ordered product> places fields with the latest time on the left. For two bosonic insertions,
$$
T\{\phi(x)\phi(y)\}=\theta(x^0-y^0)\phi(x)\phi(y)+\theta(y^0-x^0)\phi(y)\phi(x).
$$
The <normal-ordered product> $:\!\phi(x_1)\cdots\phi(x_n)\!:$ instead moves every <creation operator> to the left of every <annihilation operator>, without adding the commutators generated by that rearrangement. Its <vacuum expectation value> is zero whenever a nonempty set of free-field factors remains.
The mode <commutator> gives the positive-frequency <Wightman function>
$$
D^+(x-y)=[\phi^{(+)}(x),\phi^{(-)}(y)]
=\langle0|\phi(x)\phi(y)|0\rangle
=\int\frac{d^3p}{(2\pi)^3\,2E_{\mathbf p}}e^{-ip\cdot(x-y)}.
$$
Define the <Feynman propagator> here to be the two-point function itself, rather than that function divided by $i$. Then
$$
\boxed{\Delta_F(x-y)=\langle0|T\{\phi(x)\phi(y)\}|0\rangle
=\theta(x^0-y^0)D^+(x-y)+\theta(y^0-x^0)D^+(y-x).}
$$
Equivalently,
$$
\Delta_F(x-y)=\int\frac{d^4p}{(2\pi)^4}\frac{i\,e^{-ip\cdot(x-y)}}{p^2-m^2+i0}.
$$
The positive-energy pole is below and the negative-energy pole above the real axis. Closing the energy <contour integral> below for positive time separation and above for negative time separation gives the two terms above. Thus this is the <scalar Feynman propagator pole prescription>, and $(\Box+m^2)\Delta_F=-i\delta^4$. In particular, $T\{\phi(x)\phi(y)\}=:\!\phi(x)\phi(y)\!:+\Delta_F(x-y)$: the difference between <time ordering> and <normal ordering> is the <Wick contraction>.
For $n$ free bosonic fields, the <Wick theorem> states
$$
\boxed{T\{\phi_1\cdots\phi_n\}
=\sum_{P}\left(\prod_{(i,j)\in P}\Delta_F(x_i-x_j)\right)
:\!\prod_{r\notin V(P)}\phi_r\!:,}
$$
where $P$ runs over all sets of disjoint pairs of indices, including the empty set, and $V(P)$ is the set of paired indices. The empty <normal-ordered product> equals one. This is an operator identity for free fields, understood with smearing or a regulator when products of <operator-valued distributions> require it.
Here is an <inductive proof of the scalar Wick theorem>. Sort the arguments chronologically so that $x_1^0$ is the latest. For any remaining set $I$, commuting the annihilation part of $\phi_1$ through the creation parts in a <normal-ordered product> gives
$$
\phi_1:\!\prod_{r\in I}\phi_r\!:
=:\!\phi_1\prod_{r\in I}\phi_r\!:
+\sum_{j\in I}[\phi_1^{(+)},\phi_j^{(-)}]
:\!\prod_{r\in I\setminus\{j\}}\phi_r\!:.
$$
The creation part of $\phi_1$ is already in the correct position. Every commutator on the right is a scalar, so no further operator rearrangement is needed for it. Because $x_1$ is latest, that commutator is $\Delta_F(x_1-x_j)$. Assume the <Wick theorem> for the other $n-1$ fields and multiply its expansion by $\phi_1$. The first term of the displayed identity gives every pairing in which index 1 is unpaired; the sum gives every pairing in which 1 is paired with exactly one $j$. Each final partial pairing has exactly one such construction. The one-field identity supplies the base case, proving the theorem by <mathematical induction>. Applying <time ordering> restores arbitrary argument order.
Taking the <vacuum expectation value> retains only complete pairings. Odd free correlators vanish; for example the <free scalar four-point function> is $\Delta_{12}\Delta_{34}+\Delta_{13}\Delta_{24}+\Delta_{14}\Delta_{23}$. These facts explain why propagators are the edges of <Feynman diagrams>.
For the interacting <quartic scalar field theory>, the <interaction Hamiltonian> density is $\lambda\phi^4/4!$. With a single <time ordering> acting on external insertions and interaction vertices, the unnormalized numerator expands as
$$
N_n=\sum_{v=0}^{\infty}\frac{(-i\lambda)^v}{v!(4!)^v}
\int d^4z_1\cdots d^4z_v\,
\langle0|T\{\phi(x_1)\cdots\phi(x_n)\phi(z_1)^4\cdots\phi(z_v)^4\}|0\rangle.
$$
Apply the proved <Wick theorem> term by term. Each complete pairing connects external insertions and four-legged vertices by <Feynman propagators>. There are $4!$ assignments of four distinct incident legs at a vertex, canceling its factorial. Permuting identical integrated vertices cancels the corresponding part of $v!$. Permutations still leaving the labeled graph unchanged give its <Feynman-diagram symmetry factor>. For example,
$$
N_{4,\mathrm{connected}}^{(1)}=-i\lambda\int d^4z\prod_{r=1}^4\Delta_F(x_r-z),
$$
whereas the two-point <tadpole diagram> has coefficient $-i\lambda/2$, because choosing the two vertex legs contracted with the labeled external fields gives $4\cdot3$ possibilities and the other two contract with each other.
Thus the position-space rules attach $\Delta_F$ to every edge, $-i\lambda\int d^4z$ to every four-legged <interaction vertex>, and divide by the graph's <Feynman-diagram symmetry factor>. Fourier transformation gives the momentum-space <Feynman rules>:
$$
\boxed{\text{line: }\frac{i}{p^2-m^2+i0},\qquad
\text{vertex: }-i\lambda,\qquad
\text{loop: }\int\frac{d^4\ell}{(2\pi)^4}.}
$$
Enforce <momentum conservation> at each vertex, retaining the overall four-momentum delta function. External lines of these correlation functions remain unamputated. The unnormalized expression includes <Vacuum Feynman diagrams>; dividing by $N_0=\langle0|S|0\rangle$ cancels those disconnected vacuum factors. They should not be discarded without that normalization.
There is a time-ordering convention in the printed correlation-function notation worth making explicit. The usual interacting correlator uses the single global <time ordering> in $N_n$ above. If the displayed $T\{\phi_1\cdots\phi_n\}S$ is read literally with $S$ outside that ordering, the external block and the vertex block are separately time ordered: their cross-contractions are $D^+(x_i-z)$ because each external field stands to the left of every vertex field. The same pairing construction then uses $\Delta_F$ for lines within either block and $D^+$ for lines between blocks; in momentum space the latter is $2\pi\theta(p^0)\delta(p^2-m^2)$. This literal product is different from the conventional globally time-ordered interacting correlator, and does not have the ordinary external <Feynman propagators> derived above.
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