# _embedded.inference_rules[].latex
Indefinite integral of RHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Indefinite integral of LHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Take inner product of $#1$ with Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Expand $#1$ in Eq.~\ref{eq:#2} with conjugate; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Multiply LHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Multiply both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Select real parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Replace scalar variables in Eq.~\ref{eq:#1} with equivalent vector variables; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Judicious choice as a guessed solution to Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2},

# _embedded.inference_rules[].latex
Add zero to LHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} into Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.

# _embedded.inference_rules[].name_latex
substitute LHS of three expressions into expression

# _embedded.inference_rules[].latex
Factor $#1$ from the RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Swap LHS of Eq.~\ref{eq:#1} with RHS; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Separate two vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Subtract $#1$ from both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Separate vector in Eq.~\ref{eq:#2} into components related by angle $#1$; yields Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.

# _embedded.inference_rules[].name_latex
separate vector into two trigonometric ratios

# _embedded.inference_rules[].latex
$#1$ is even with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.

# _embedded.inference_rules[].latex
Eq.~\ref{eq:#1} is one of the final equations.

# _embedded.inference_rules[].latex
Indefinite integral of both sides of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Thus we see that LHS of Eq.~\ref{eq:#1} is equal to RHS.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} into Eq.~\ref{eq:#6}; yields Eq.~\ref{eq:#7}.

# _embedded.inference_rules[].name_latex
substitute LHS of five expressions into expression

# _embedded.inference_rules[].latex
LHS of Eq.~\ref{eq:#1} is equal to LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Take cross product of Eq.~\ref{eq:#2} and $#1$; yields Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Sum exponents on LHS and RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} and LHS of Eq.~\ref{eq:#6} into Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}.

# _embedded.inference_rules[].name_latex
substitute LHS of six expressions into expression

# _embedded.inference_rules[].latex
Apply operator in Eq.~\ref{eq:#1} to bra; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].notes_latex
not clear how to specify limits for an arbitrary number of integrals

# _embedded.inference_rules[].latex
Sum exponents on RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Apply divergence to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Take cross product of $#1$ and Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Distribute conjugate to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Conjugate both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Apply function $#1$ with argument $#2$ to Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}

# _embedded.inference_rules[].name_latex
apply function to both sides of expression

# _embedded.inference_rules[].latex
$#1$ is odd with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.

# _embedded.inference_rules[].latex
Take the square root of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Add $#1$ to both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Apply gradient to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Expand the RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Thus we see that Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Separate three vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Expand the LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Sum exponents on LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Take inner product of Eq.~\ref{eq:#2} with $#1$; yields Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Expand integrand of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Indefinite integral of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Factor $#1$ from the LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Multiply RHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Partially differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Conjugate transpose of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Eq.~\ref{eq:#1} is valid when Eq.~\ref{eq:#2} occurs; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Replace constant $#1$ with value $#2$ and units $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}

# _embedded.inference_rules[].latex
Add zero to RHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Make Eq.~\ref{eq:#2} the power of $#1$; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Combine like terms in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} into Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.

# _embedded.inference_rules[].name_latex
substitute LHS of two expressions into expression

# _embedded.inference_rules[].latex
Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2} under the condition in Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].name_latex
expr 1 is equivalent to expr 2 under the condition

# _embedded.inference_rules[].latex
The maximum of Eq.~\ref{eq:#2} with respect to $#1$ is Eq.~\ref{eq:#3}

# _embedded.inference_rules[].latex
Apply curl to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Based on the assumption $#1$, drop non-dominant term in Eq.~\ref{#2}; yeilds Eq.~\ref{#3}

# _embedded.inference_rules[].latex
Assume $#1$ dimensions; decompose vector to be Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} into Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}.

# _embedded.inference_rules[].name_latex
substitute LHS of four expressions into expression

# _embedded.inference_rules[].latex
A boundary condition for Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2}

# _embedded.inference_rules[].latex
RHS of Eq.~\ref{eq:#1} is equal to RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.

# _embedded.inference_rules[].latex
Distribute conjugate transpose to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].name_latex
distribute conjugate transpose to factors

# _embedded.inference_rules[].latex
Select imaginary parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Replace summation notation in Eq.~\ref{eq:#1} with vector notation; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].name_latex
replace summation notation with vector notation

# _embedded.inference_rules[].latex
Replace curl in Eq.~\ref{eq:#1} with Levi-Cevita contravariant; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].name_latex
replace curl with LeviCevita summation contravariant

# _embedded.inference_rules[].latex
Integrate Eq.~\ref{eq:#4} over $#1$ from lower limit $#2$ to upper limit $#3$; yields Eq.~\ref{eq:#5}.

# _embedded.inference_rules[].latex
Apply operator in Eq.~\ref{eq:#1} to ket; yields Eq.~\ref{eq:#2}.

# _embedded.inference_rules[].latex
Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.