Physics Derivation Graph navigation Sign in

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. 41
declare guess solution 1 1 0 Judicious choice as a guessed solution to Eq.~ref{eq:#1} is 2
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. 110
differentiate with respect to 1 1 1 Differentiate Eq.~ref{eq:#2} with respect to $#1$; yields Eq.~ref{eq:#3}. 3 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}. 36
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}. 30
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}. 69
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}. 38
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}. 13
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}. 5
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

Add new inference rule

This action requires you to be signed in

Delete an inference rule

This action requires you to be signed in

Rename an inference rule

This action requires you to be signed in

Edit the latex for an inference rule

This action requires you to be signed in