Problem $4$: Odd and even integers

# Return the following $6$ laws. Be sure your formulas are exactly in the order specified.

#$0.$ Each number $$$$ has exactly one successor, which is not equal to $$$$$.$

# $1.$ Each number is either even or odd, but not both.

# $2.$ The successor number of an even number is odd.

# $3.$ The successor number of an odd number is even.

# $4.$ For every number $$x$$$,$ the successor of $$x$$ is larger than $$x$$$.$

# $5.$ Larger is a transitive property: if $$x$$ is larger than $$y$$ and $$y$$ is

# larger than $$z$$$,$ then $$x$$ is larger than $$z$$$.$

# Query: For each number, there exists an even number larger than it$.$

def ints$()->$ Tuple$[$List$[$Formula$],$ Formula$]$:

def Even $(x)$ : return Atom$(\text{'}$Even$\text{'},x)$ # whether $x$ is even

def Odd$(x)$ : return Atom$(\text{'}$Odd$\text{'},x)$ # whether $x$ is odd

def Successor$(x,y)$: return Atom$(\text{'}$Successor$\text{'},x,y)$ # whether x$\text{'}$s successor is $y$

def Larger$(x,y$ : return Atom$(\text{'}$Larger$\text{'},x,y$ whether $x$ is larger than $y$

# Note: all objects are numbers, so we don't need to define Number as an

# explicit predicate.

# Note: pay attention to the order of arguments of Successor and Larger.

# Populate $|$formulas$|$ with the $6$ laws above and set $|$query$|$ to be the

# query.

# Hint: You might want to use the Equals predicate, defined in

logic.py$.$ This

# predicate is used to assert that two objects are the same.

formulas $=[]$

query $=$ None

# BEGIN$\_$YOUR$\_$CODE $($our solution is $23$ lines of code, but don't worry if you deviate from this$)$

formulas.append$($Exists$())$

# END$\_$YOUR$\_$CODE

query ${?}^{-}=$Forall$(\text{'}$$x$\text{'},$ Exists$(\text{'}$$y$\text{'},$ And$($Even$(\text{'}$$y$\text{'}),$ Larger$(\text{'}$$y$\text{'},\text{'}$$x$\text{'}))))$

return $($formulas$,$ query$)$