## Inference rules in the Physics Derivation Graph

See the documentation for background on inference rules.

Name:

inf rule name inputs outputs feeds Latex Used in derivations number of uses assumptions
apply operator to bra 1 1 0 Apply operator in Eq.~ref{eq:#1} to bra; yields Eq.~ref{eq:#2}. 1 quantum: Dirac notation
apply operator to ket 1 1 0 Apply operator in Eq.~ref{eq:#1} to ket; yields Eq.~ref{eq:#2}. 1 quantum: Dirac notation
X cross both sides by 1 1 1 Take cross product of $#1$ and Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}.
0 linear algebra
X dot both sides 1 1 1 Take inner product of $#1$ with Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}.
0 linear algebra
add X to both sides 1 1 1 Add $#1$ to both sides of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 15
add zero to LHS 1 1 1 Add zero to LHS of Eq.~ref{eq:#2}, where $0=#1$; yields Eq.~ref{eq:#3}.
0
add zero to RHS 1 1 1 Add zero to RHS of Eq.~ref{eq:#2}, where $0=#1$; yields Eq.~ref{eq:#3}.
0
apply divergence 1 1 0 Apply divergence to both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 multi-variable calculus
apply gradient to scalar function 1 1 0 Apply gradient to both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 multi-variable calculus
assume N dimensions 0 1 1 Assume $#1$ dimensions; decompose vector to be Eq.~ref{eq:#2}. 3
both sides cross X 1 1 1 Take cross product of Eq.~ref{eq:#2} and $#1$; yields Eq.~ref{eq:#3}
0 linear algebra
both sides dot X 1 1 1 Take inner product of Eq.~ref{eq:#2} with $#1$; yields Eq.~ref{eq:#3}
0 linear algebra
boundary condition for expr 1 1 0 A boundary condition for Eq.~ref{eq:#1} is Eq.~ref{eq:#2} 2
claim LHS equals RHS 1 0 0 Thus we see that LHS of Eq.~ref{eq:#1} is equal to RHS. 3
combine like terms 1 1 0 Combine like terms in Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1
conjugate both sides 1 1 0 Conjugate both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 3 complex values
conjugate function X 1 1 1 Conjugate $#1$ in Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 1 complex values
conjugate transpose both sides 1 1 0 Conjugate transpose of both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 complex valued linear algebra
declare assumption 0 1 0 Eq.~ref{eq:#1} is an assumption. 14
declare final expr 1 0 0 Eq.~ref{eq:#1} is one of the final equations. 42
declare guess solution 1 1 0 Judicious choice as a guessed solution to Eq.~ref{eq:#1} is Eq.~ref{eq:#2}, 3
declare identity 0 1 0 Eq.~ref{eq:#1} is an identity. 12
declare initial expr 0 1 0 Eq.~ref{eq:#1} is an initial equation. 113
differentiate with respect to 1 1 1 Differentiate Eq.~ref{eq:#2} with respect to $#1$; yields Eq.~ref{eq:#3}. 5 differential equations
distribute conjugate to factors 1 1 0 Distribute conjugate to factors in Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 complex values
distribute conjugate transpose to factors 1 1 0 Distribute conjugate transpose to factors in Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 complex valued linear algebra
divide both sides by 1 1 1 Divide both sides of Eq.~ref{eq:#2} by $#1$; yields Eq.~ref{eq:#3}. 37
evaluate definite integral 1 1 0 Evaluate definite integral Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 3 multi-variable calculus
expand LHS 1 1 0 Expand the LHS of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.
0
expand RHS 1 1 0 Expand the RHS of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1
expand integrand 1 1 0 Expand integrand of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 multi-variable calculus
expand magnitude to conjugate 1 1 1 Expand $#1$ in Eq.~ref{eq:#2} with conjugate; yields Eq.~ref{eq:#3}. 1 complex values
factor out X 1 1 1 Factor $#1$ from Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}.
0
factor out X from LHS 1 1 1 Factor $#1$ from the LHS of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 2
factor out X from RHS 1 1 1 Factor $#1$ from the RHS of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 1
function is even 1 1 3 $#1$ is even with respect to $#2$, so replace $#1$ with $#3$ in Eq.~ref{eq:#4}; yields Eq.~ref{eq:#5}. 1
function is odd 1 1 3 $#1$ is odd with respect to $#2$, so replace $#1$ with $#3$ in Eq.~ref{eq:#4}; yields Eq.~ref{eq:#5}. 1
indefinite integral over 1 1 1 Indefinite integral of both sides of Eq.~ref{eq:#2} over $#1$; yields Eq.~ref{eq:#3}.
0 multi-variable calculus
indefinite integrate LHS over 1 1 1 Indefinite integral of LHS of Eq.~ref{eq:#2} over $#1$; yields Eq.~ref{eq:#3}.
0 multi-variable calculus
indefinite integration 1 1 0 Indefinite integral of both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 5 multi-variable calculus
indefinite integrate RHS over 1 1 1 Indefinite integral of RHS of Eq.~ref{eq:#2} over $#1$; yields Eq.~ref{eq:#3}. 2 multi-variable calculus
integrate over from to 1 1 3 Integrate Eq.~ref{eq:#4} over $#1$ from lower limit $#2$ to upper limit $#3$; yields Eq.~ref{eq:#5}.
0 multi-variable calculus
make expr power 1 1 1 Make Eq.~ref{eq:#2} the power of $#1$; yields Eq.~ref{eq:#3}. 1
multiply LHS by unity 1 1 1 Multiply LHS of Eq.~ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~ref{eq:#3}
0
multiply RHS by unity 1 1 1 Multiply RHS of Eq.~ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~ref{eq:#3} 2
multiply both sides by 1 1 1 Multiply both sides of Eq.~ref{eq:#2} by $#1$; yields Eq.~ref{eq:#3}. 31
normalization condition 0 1 0 Normalization condition is Eq.~ref{eq:#1}. 1
partially differentiate with respect to 1 1 1 Partially differentiate Eq.~ref{eq:#2} with respect to $#1$; yields Eq.~ref{eq:#3}. 2 differential equations
raise both sides to power 1 1 1 Raise both sides of Eq.~ref{eq:#2} to $#1$; yields Eq.~ref{eq:#3}. 5
replace curl with LeviCevita summation contravariant 1 1 0 Replace curl in Eq.~ref{eq:#1} with Levi-Cevita contravariant; yields Eq.~ref{eq:#2}. 2 linear algebra
replace scalar with vector 1 1 0 Replace scalar variables in Eq.~ref{eq:#1} with equivalent vector variables; yields Eq.~ref{eq:#2}. 1
replace summation notation with vector notation 1 1 0 Replace summation notation in Eq.~ref{eq:#1} with vector notation; yields Eq.~ref{eq:#2}. 1 linear algebra
select imaginary parts 1 1 0 Select imaginary parts of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.
0 complex values
select real parts 1 1 0 Select real parts of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 complex values
simplify 1 1 0 Simplify Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 66
solve for X 1 1 1 Solve Eq.~ref{eq:#2} for $#1$; yields Eq.~ref{eq:#3}.
0
change variable X to Y 1 1 2 Change variable $#1$ to $#2$ in Eq.~ref{eq:#3}; yields Eq.~ref{eq:#4}. 24
subtract X from both sides 1 1 1 Subtract $#1$ from both sides of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 16
sum exponents 1 1 0 Sum exponents on LHS and RHS of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.
0
sum exponents LHS 1 1 0 Sum exponents on LHS of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.
0
sum exponents RHS 1 1 0 Sum exponents on RHS of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.
0
swap LHS with RHS 1 1 0 Swap LHS of Eq.~ref{eq:#1} with RHS; yields Eq.~ref{eq:#2}. 14
take curl of both sides 1 1 0 Apply curl to both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 1 linear algebra
multiply expr 1 by expr 2 2 1 0 Multiply Eq.~ref{eq:#1} by Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 8
substitute LHS of expr 1 into expr 2 2 1 0 Substitute LHS of Eq.~ref{eq:#1} into Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 70
substitute RHS of expr 1 into expr 2 2 1 0 Substitute RHS of Eq.~ref{eq:#1} into Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 39
LHS of expr 1 equals LHS of expr 2 2 1 0 LHS of Eq.~ref{eq:#1} is equal to LHS of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 16
RHS of expr 1 equals RHS of expr 2 2 1 0 RHS of Eq.~ref{eq:#1} is equal to RHS of Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 1
subtract expr 1 from expr 2 2 1 0 Subtract Eq.~ref{eq:#1} from Eq.~ref{eq:#2}; yields Eq.~ref{eq:#2}. 4
add expr 1 to expr 2 2 1 0 Add Eq.~ref{eq:#1} to Eq.~ref{eq:#2}; yields Eq.~ref{eq:#2}. 3
expr 1 is true under condition expr 2 2 1 0 Eq.~ref{eq:#1} is valid when Eq.~ref{eq:#2} occurs; yields Eq.~ref{eq:#3}. 1
claim expr 1 equals expr 2 2 0 0 Thus we see that Eq.~ref{eq:#1} is equivalent to Eq.~ref{eq:#2}.
0
apply function to both sides of expression 1 1 2 Apply function $#1$ with argument $#2$ to Eq.~ref{eq:#3}; yields Eq.~ref{eq:#4} 1
expr 1 is equivalent to expr 2 under the condition 2 1 0 Eq.~ref{eq:#1} is equivalent to Eq.~ref{eq:#2} under the condition in Eq.~ref{eq:#3}. 3
substitute LHS of two expressions into expr 3 1 0 Substitute LHS of Eq.~ref{eq:#1} and LHS of Eq.~ref{eq:#2} into Eq.~ref{eq:#3}; yields Eq.~ref{eq:#4}. 6
substitute LHS of three expressions into expr 4 1 0 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}. 2
substitute LHS of four expressions into expr 5 1 0 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}. 2
substitute LHS of five expressions into expr 6 1 0 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}.
0
substitute LHS of six expressions into expr 7 1 0 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}.
0
change two variables in expr 1 1 4 Change variable $#1$ to $#2$ and $#3$ to $#4$ in Eq.~ref{eq:#5}; yields Eq.~ref{eq:#6}. 21
change three variables in expr 1 1 6 Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ in Eq.~ref{eq:#7}; yields Eq.~ref{eq:#8}. 7
change four variables in expr 1 1 8 Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ in Eq.~ref{eq:#9}; yields Eq.~ref{eq:#10}. 2
change five variables in expr 1 1 10 Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ and $#9$ to $#10$ in Eq.~ref{eq:#11}; yields Eq.~ref{eq:#12}.
0
change six variables in expr 1 1 12 Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ and $#9$ to $#10$ and $#11$ to $#12$ in Eq.~ref{eq:#13}; yields Eq.~ref{eq:#14}.
0
square root both sides 1 2 0 Take the square root of both sides of Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2} and Eq.~ref{eq:#3}. 6
divide expr 1 by expr 2 2 1 0 Divide Eq.~ref{eq:#1} by Eq.~ref{eq:#2}; yields Eq.~ref{eq:#3}. 1
separate two vector components 1 2 0 Separate two vector components in Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2} and Eq.~ref{eq:#3} 1
separate three vector components 1 3 0 Separate three vector components in Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2} and Eq.~ref{eq:#3} and Eq.~ref{eq:#4}.
0
separate vector into two trigonometric ratios 1 2 1 Separate vector in Eq.~ref{eq:#2} into components related by angle $#1$; yields Eq.~ref{eq:#3} and Eq.~ref{eq:#4}. 1
boundary condition 1 1 0 Boundary condition: Eq.~ref{eq:#2} when Eq.~ref{eq#1}. 2
maximum of expr 1 1 1 The maximum of Eq.~ref{eq:#2} with respect to $#1$ is Eq.~ref{eq:#3} 2
integrate 1 1 0 Integrate Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. 2 multi-variable calculus
replace constant with value 1 1 3 Replace constant $#1$ with value $#2$ and units $#3$ in Eq.~ref{eq:#4}; yields Eq.~ref{eq:#5} 3
drop non-dominant term 1 1 1 Based on the assumption $#1$, drop non-dominant term in Eq.~ref{#2}; yeilds Eq.~ref{#3} 2
Physics Derivation Graph: 94 Inference Rules

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