[proofplan] Fix $i$, $b$, and $c$, and write the separately reduced $i$-signatures of $b$ and $c$ in terms of their surviving plus and minus signs. Under the stated [tensor product](/page/Tensor%20Product) convention, the only cancellations that can newly occur after concatenation are cross-boundary cancellations between surviving minus signs from $b$ and surviving plus signs from $c$. Counting these cancellations compares exactly $\varepsilon_i(b)$ with $\varphi_i(c)$, and the tensor product rule follows by locating the rightmost surviving $+$ for $f_i$ and the leftmost surviving $-$ for $e_i$. [/proofplan] [step:Write the separately reduced signatures using $\varepsilon_i$ and $\varphi_i$] Fix $i\in\{1,\dots,n-1\}$, $b\in B$, and $c\in C$. By the definition of the reduced $i$-signature and the string statistics, the reduced $i$-signature of $b$ consists of $\varphi_i^B(b)$ surviving plus signs followed by $\varepsilon_i^B(b)$ surviving minus signs, and the reduced $i$-signature of $c$ consists of $\varphi_i^C(c)$ surviving plus signs followed by $\varepsilon_i^C(c)$ surviving minus signs. Define \begin{align*} p_b:=\varphi_i^B(b),\qquad q_b:=\varepsilon_i^B(b),\qquad p_c:=\varphi_i^C(c),\qquad q_c:=\varepsilon_i^C(c). \end{align*} These are finite nonnegative integers because the $i$-strings in $B$ and $C$ are finite. Thus the concatenated reduced signature of $b\otimes c$ has the form \begin{align*} +^{p_b}-^{q_b}+^{p_c}-^{q_c}, \end{align*} where $+^r$ denotes a block of $r$ plus signs and $-^r$ denotes a block of $r$ minus signs. [/step] [step:Identify the only possible cross-boundary cancellations] Inside each factor, the signature has already been reduced, so no adjacent cancellable pair $-+$ remains within the block $+^{p_b}-^{q_b}$ or within the block $+^{p_c}-^{q_c}$. Therefore any new adjacent pair $-+$ in the concatenated signature must use a minus sign from the final block $-^{q_b}$ of the $b$-signature and a plus sign from the initial block $+^{p_c}$ of the $c$-signature. The cancellation rule removes as many such cross-boundary pairs as possible. Hence the number of cross-boundary cancellations is \begin{align*} r:=\min\{q_b,p_c\}. \end{align*} After these cancellations, the surviving signs from the middle two blocks are precisely $p_c-r$ plus signs from $c$ and $q_b-r$ minus signs from $b$. [guided] The separately reduced signatures contain no internal $-+$ pairs, so after we concatenate them the only place a new cancellable pair can occur is at the boundary between the two factors. More explicitly, the reduced signature of $b$ has shape $+^{p_b}-^{q_b}$ and the reduced signature of $c$ has shape $+^{p_c}-^{q_c}$. Concatenating in the convention of the theorem gives \begin{align*} +^{p_b}-^{q_b}+^{p_c}-^{q_c}. \end{align*} The first block $+^{p_b}$ cannot contribute the minus sign in a pair $-+$. The last block $-^{q_c}$ cannot contribute the plus sign in such a pair. Inside $+^{p_b}-^{q_b}$ and inside $+^{p_c}-^{q_c}$ there is also no $-+$ pair, because each factor was already reduced. Thus every new cancellation pairs one of the $q_b$ surviving minus signs from $b$ with one of the $p_c$ surviving plus signs from $c$. The cancellation rule removes adjacent $-+$ pairs until no such pair remains. Therefore the number of cross-boundary cancellations is exactly \begin{align*} r=\min\{q_b,p_c\}. \end{align*} If $p_c>q_b$, then all $q_b$ middle minus signs from $b$ are cancelled and $p_c-q_b$ plus signs from $c$ survive. If $p_c\le q_b$, then all $p_c$ middle plus signs from $c$ are cancelled and $q_b-p_c$ minus signs from $b$ survive. This is the whole source of the strict and weak inequalities in the tensor product rule. [/guided] [/step] [step:Locate the selected plus sign for $f_i$] By the tensor product signature rule, $f_i$ acts on the tensor factor containing the rightmost surviving plus sign in the reduced concatenated signature. Suppose first that $\varphi_i^C(c)>\varepsilon_i^B(b)$. In the notation above, this says $p_c>q_b$. Then $r=q_b$, so $p_c-r=p_c-q_b>0$. Thus at least one plus sign from the $c$-signature survives all cross-boundary cancellations. Every surviving plus sign from $c$ lies to the right of every sign from the $b$-signature, so the rightmost surviving plus sign lies in the $c$-factor. Therefore, whenever $f_i^C(c)$ is defined, \begin{align*} f_i^{B\otimes C}(b\otimes c)=b\otimes f_i^C(c). \end{align*} Suppose next that $\varphi_i^C(c)\le\varepsilon_i^B(b)$. In the notation above, this says $p_c\le q_b$. Then $r=p_c$, so every plus sign from the $c$-signature is cancelled. Consequently, if a surviving plus sign remains in the reduced concatenated signature, it must come from the $b$-signature. If no plus sign survives, then the tensor product operator is $0$ and the asserted formula is outside the stated defined case. Therefore, whenever $f_i^B(b)$ is defined, \begin{align*} f_i^{B\otimes C}(b\otimes c)=f_i^B(b)\otimes c. \end{align*} [/step] [step:Locate the selected minus sign for $e_i$] By the tensor product signature rule, $e_i$ acts on the tensor factor containing the leftmost surviving minus sign in the reduced concatenated signature. Suppose first that $\varepsilon_i^B(b)>\varphi_i^C(c)$. In the notation above, this says $q_b>p_c$. Then $r=p_c$, so $q_b-r=q_b-p_c>0$. Thus at least one minus sign from the $b$-signature survives all cross-boundary cancellations. Every surviving minus sign from $b$ lies to the left of every sign from the $c$-signature, so the leftmost surviving minus sign lies in the $b$-factor. Therefore, whenever $e_i^B(b)$ is defined, \begin{align*} e_i^{B\otimes C}(b\otimes c)=e_i^B(b)\otimes c. \end{align*} Suppose next that $\varepsilon_i^B(b)\le\varphi_i^C(c)$. In the notation above, this says $q_b\le p_c$. Then $r=q_b$, so every minus sign from the $b$-signature is cancelled. Consequently, if a surviving minus sign remains in the reduced concatenated signature, it must come from the $c$-signature. If no minus sign survives, then the tensor product operator is $0$ and the asserted formula is outside the stated defined case. Therefore, whenever $e_i^C(c)$ is defined, \begin{align*} e_i^{B\otimes C}(b\otimes c)=b\otimes e_i^C(c). \end{align*} [/step] [step:Translate the sign locations into the stated tensor product formulas] The preceding two steps determine exactly which tensor factor contains the sign selected by the tensor product signature rule. Replacing $p_c$ by $\varphi_i^C(c)$ and $q_b$ by $\varepsilon_i^B(b)$ gives the following two formulas for $f_i^{B\otimes C}$ whenever the indicated factor operator is defined. If $\varphi_i^C(c)>\varepsilon_i^B(b)$, then \begin{align*} f_i^{B\otimes C}(b\otimes c)=b\otimes f_i^C(c). \end{align*} If $\varphi_i^C(c)\le \varepsilon_i^B(b)$, then \begin{align*} f_i^{B\otimes C}(b\otimes c)=f_i^B(b)\otimes c. \end{align*} Likewise, the two formulas for $e_i^{B\otimes C}$ are as follows whenever the indicated factor operator is defined. If $\varepsilon_i^B(b)>\varphi_i^C(c)$, then \begin{align*} e_i^{B\otimes C}(b\otimes c)=e_i^B(b)\otimes c. \end{align*} If $\varepsilon_i^B(b)\le \varphi_i^C(c)$, then \begin{align*} e_i^{B\otimes C}(b\otimes c)=b\otimes e_i^C(c). \end{align*} These are precisely the asserted tensor product rules. [/step]