Show that minimal CFG is undecidable via mapping reduction
Here is a solution based on a reduction from Post Correspondence Problem.  The general idea is similar to the ones used in reductions from PCP to the emptiness of intersection and ambiguity in context-free grammars.  I am afraid I am using a Turing reduction for a preliminary claim---which I would consider fair game considering the number of variants of PCP `out there'.

Preliminaries. 
A PCP instance is a finite set of pairs $\Pi=(u_i,v_i)_{i&lt;n}$ of finite words over a finite alphabet $\Sigma$.  A solution is a sequence of indices $j_0,\dots,j_m$ with $u_{j_0}\cdots u_{j_m}=v_{j_0}\cdots v_{j_m}$.  The set of indices $I(\Pi)$ of a PCP instance $\Pi$ is the set of indices $i&lt;n$ that appear in some solution; formally $$I(\Pi)=\{ i &lt; n\mid \exists i_0,\dots,i_m &lt;n\mathbin.u_{j_0}\cdots u_{j_m}=v_{j_0}\cdots v_{j_m}\text{ and }\exists 0\leq k\leq m\mathbin.i=j_k\}\;.$$
Then the PCP instance has a solution iff $I(\Pi)\neq\emptyset$.

Claim.  The following decision problem called full PCP is also undecidable:


instance $\Pi$
question $I(\Pi)=\{0,\dots,n-1\}$?


Proof.  By a Turing reduction from PCP.  Assume full PCP is decidable, and consider a PCP instance $\Pi$.  For a subset $J\subseteq\{0,\dots,n-1\}$, let $\Pi_J=(u_i,v_i)_{i\in J}$.  Then $\Pi$ has a solution iff $I(\Pi)\neq\emptyset$ iff there exists $J\subseteq\{0,\dots,n-1\}$ such that $\Pi_J$ is a positive instance of the full PCP.  Thus it suffices to check all the subsets $J$ with the algorithm for the full PCP; if at least one succeeds, the original PCP instance was positive, otherwise it was negative. $\Box$

The first grammar $G_1$ is constructed from a full PCP instance $\Pi$ in order to recognise
$$\begin{aligned}L_1&amp;=\{ \$u_{j_0}\cdots \$u_{j_m}\#(\$v_{j_0}\cdots \$v_{j_m})^R\mid j_0,\dots,j_m &lt; n\}\;,\\L'_1&amp;=L_1\cup\{\#\}\end{aligned}$$
where $\cdot^R$ denotes the reversal operation, and $\$,\#$ are fresh symbols not in $\Sigma$.  Indeed, we can define $G_1=(\{S_1\},\Sigma\uplus\{\$,\#\},P_1,S_1)$ with $$P_1=\{S_1\to \$u_j S_1 v_j^R\$\mid j&lt;n\}\cup\{S_1\to\#\}.$$ Then $L'_1=L(G_1)$.
Without loss of generality, we can assume $(u_i,v_i)\neq(u_j,v_j)$ for all $i\neq j&lt;n$, and the grammar $G_1$ is minimal: the $\$$ symbols are indeed here to ensure minimality (as otherwise there might exist two different sequences of indices yielding the same words).

The second grammar generates `centered non-palindromes'
$$L_2=\{w\#{w'}^R\mid w\neq w'\in\Sigma^*\}.$$  Note that a word $w\#{w'}^R$ in $L_2$ is such that $w$ and $w'$ share a longest common prefix $u$: $w=uv$ and $w'=uv'$ for some $v,v'\in\Sigma^*$ with


either $v=ax$ and $v'=bx'$ for some $a\neq b\in\Sigma$ and $x,x'\in\Sigma^\ast$,
or $v=ax$ and $v'=\varepsilon$ for some $a\in\Sigma$ and $x\in\Sigma^\ast$,
or $v=\varepsilon$ and $v'=ax'$ for some $a\in\Sigma$ and $x'\in\Sigma^\ast$.


Thus $L_2$ is generated by $G_2=(\{S_2,E,A\},\Sigma\uplus\{\#\},P_2,S_2)$ with $$\begin{aligned}P_2&amp;=\{S_2\to aS_2a\mid a\in\Sigma\}\cup\{S_2\to E\}\\&amp;\cup\{E\to aA\#Ab\mid a\neq b\in\Sigma\}\cup\{aA\#\mid a\in\Sigma\}\cup\{\#Aa\mid a\in\Sigma\}\\&amp;\cup\{A\to aA\mid a\in\Sigma\}\cup\{A\to\varepsilon\}\end{aligned}$$
The idea is that $S_2$ generates $u E u^R$ for $u$ the longest common prefix of $w$ and $w'$, after which $E$ applies one of the three cases above.  Thus $L(G_2)=L_2$, and the grammar is minimal.

Relating the two grammars.  Let $h$ be the homomorphism from $(\Sigma\uplus\{\$,\#\})^*$ to $(\Sigma\uplus\{\#\})^*$ defined by $h(a)=a$ for all $a\in\Sigma$, $h(\#)=\#$, and $h(\$)=\varepsilon$; $h$ simply erases $\$$ symbols.  Then $\Pi$ has a solution iff $h(L_1)\not\subseteq L_2$.  We define $L'_2=h^{-1}(L_2)$ and construct the corresponding grammar $G'_2=(N'_2,\Sigma\cup\{\$,\#\},P'_2,S'_2)$ in the obvious way (add a nonterminal $D$ with productions $D\to \$D$ and $D\to\varepsilon$ and introduce $D$ symbols everywhere in the right hand sides of the productions in $P_2$); it is still minimal and $\Pi$ has a solution iff $L_1\not\subseteq L'_2$.

We are however interested in the full PCP on $\Pi$.  Construct $$G=(N_1\uplus N'_2\uplus\{S\},\Sigma\uplus\{\$,\#\},P_1\cup P'_2\cup\{S\to S_1,S\to S'_2\},S) .$$

If $\Pi$ is a positive instance of the full PCP, then it has solutions, which describe words $\$u_{j_0}\cdots \$u_{j_m}\#(\$v_{j_0}\cdots \$v_{j_m})^R\not\in L'_2$ and moreover require every production of $G_1$.  Regarding $G'_2$, it suffices to consider words from $L_2\subseteq L'_2$ with no occurrence of $\$$, which do not belong to $L_1$ and therefore all the productions of $G'_2$ are needed.  Therefore, $G$ is minimal.
Otherwise, there exists $i&lt;n$ such that $i\not\in I(\Pi)$, and the production $S_1\to \$u_i S_1 v_i^R\$$ can be removed, since it cannot be part of a solution of $\Pi$, and therefore any generated word that uses this production is in $L(G'_2)$; thus $G$ is not minimal.  In particular, if $\Pi$ has no solution ($I(\Pi)=\emptyset$), then all the productions in $P_1$ except $S_1\to\#$ can be removed. $\Box$