Subtracting using minus function in alloy - subtraction

Question

I have created a vending machine it works fine. I want to subtract 1 from the quantity of items once the transaction is completed. I have provided comments in my code for understanding. Ignore some of the comments in pred chocolate. Somehow I am trying to subtract but it just wont. I don't know what seems to be the problem. Any help is appreciated .
sig button {
qty:Int} // buttons on vending machine for selecting chocolates
//sig coin{}
sig choco{
value:Int, // Each chocolate has some cost as an attribute aka value.
choice :one button // Selecting option
}
fact {
// all c:choco | all c1:choco -c | c1.choice != c.choice
}
sig machine{
cust : one customer, // Customer entity
a,b,c,d,nullb ,ip: one button, // buttons on vending machine ,ip is the input selected by user
//oners,twors,fivers ,tenrs,null1: set coin,
ipp,opc2 : one Coin, // ipp = input rs , opc = balance rs
customeripb: cust one -> one ip, // customer presses a button
customeripc: cust one -> one ipp, // customer enters coins
customeropc: opc2 one -> one cust, //customer receives balance of coins
op: one choco , // output of chocolate from machine
customerop: op one -> one cust, // customer receives a chocolate
cadbury, kitkat, eclairs , gum,null: lone choco // types of chocolate
}
{
//#(fivers+tenrs+null+twors+oners) = 5
#(a+b+c+d) = 4 // no of buttons of a b c and d are 4 on machine
# (cadbury+kitkat+ eclairs +gum) =4 // no of options to choose = 4
cadbury=choice.a // cadbury corresponds to button a
cadbury.value= 10 // cadbury costs 10rs
kitkat=choice.b // kitkat corresponds to button b
kitkat.value=5 // kitkat costs 5rs
null.value=0 // null costs 0 rs
null=choice.nullb
// as such null doesnt exist it is just to specify no i/p no o/p and nulb is an imaginary button
eclairs=choice.c // eclairs corresponds to button c
eclairs.value=1 // eclairs costs 1 rs
gum=choice.d // gum corresponds to button d
gum.value=2 // gum costs 1 rs
a.qty>=10 and a.qty<=40
b.qty>=11 and b.qty<=40
c.qty>=12 and c.qty<=40
d.qty>=13 and d.qty<=40
nullb.qty=0
//ip=nullb //input button selection is never nullb(which is imaginary button)
ipp.value!=0 // input of coins is never = 0rs
/* all m:machine|all o:opc2
|all opp: op| all i:ip|all ii:ipp| all c:m.cust
|c -> i in m.customeripb and c->ii in m.customeripc and o->c in m.customerop and opp->c in m.customerop
*/
//button=!=none
}
sig customer //user of machine
{
}
abstract sig Coin { //each coin has a valueof rs
value: Int
}
sig Nullrs extends Coin {} { value = 0 } // void rs
sig Oners extends Coin {} { value = 1 } // one rs
sig Twors extends Coin {} { value = 2 } // twors
sig Fivers extends Coin {}{ value = 5 } // five rs
sig Tenrs extends Coin {} { value = 10 } // ten rs
sig Threers extends Coin {} { value = 3 } // this is only used in o/p to specify 3rs will come out
sig Fourrs extends Coin {} { value = 4 }// this is only used in o/p to specify 4rs will come out
sig Sixrs extends Coin {} { value = 6 }// this is only used in o/p to specify 6rs will come out
sig Sevenrs extends Coin {}{ value = 7 }// this is only used in o/p to specify 7rs will come out
sig Eightrs extends Coin {} { value = 8 } // this is only used in o/p to specify 8rs will come out
sig Niners extends Coin {} { value = 9} //// this is only used in o/p to specify 9rs will come out
pred show{} // show
pred chocolate [before,after:machine ] // machine has two states one before o/p and one after
{
before.cadbury=after.cadbury
before.kitkat=after.kitkat
before.eclairs=after.eclairs
before.gum=after.gum
//all chocolates will not change and are fixed
before.ipp.value=after.ipp.value
// input value of rs remains same i.e i/p is inside machine once inputed so it cant change
before.opc2.value=0 // before state o/p value of balance coins =0
before.op=before.null // beforestate o/p = no chocolate
before.ip!=before.nullb // input button pressed never equals nullb
after.ip!=after.nullb // input button pressed never equals nullb
//before.ip=after.ip // input button pressed remains same
after.op=after.kitkat or after.op=after.eclairs
before.null=after.null // imaginary null chocolate remains in same state
before.opc2!=none and after.opc2 !=none
// balance of coins is never empty in case of empty I have defined nullrs
(after.op.value=before.ipp.value=>after.opc2.value=0)
//
(after.op=after.null=>after.opc2.value=before.ipp.value)
(before.ipp.value > after.op.value=>after.opc2.value=before.ipp.value-after.op.value)
//(before.ipp.value=after.op.value=>after.opc2.value=0)
//opc2.value!=ipp.value
before.ip=before.a or before.ip=before.b or before.ip=before.c or before.ip=before.d
(after.op=after.cadbury ) => ( ( after.ip=after.a and after.a.qty=minus[before.a.qty,1])) else
(after.op=after.kitkat ) => ( (after.ip=after.b and after.b.qty=minus[before.b.qty, 1])) else
(after.op=after.eclairs ) =>( (after.ip=after.c and after.c.qty=minus[before.c.qty,1])) else
(after.op=after.gum ) =>((after.ip=after.d and after.d.qty=minus[before.d.qty,1])) else
(after.ip=before.ip and after.ip.qty=minus[before.ip.qty,0] )
after.op!=before.null => after.op.choice=before.ip
(after.op=before.gum=>before.ipp.value>=Twors.value)
after.op=before.cadbury=>before.ipp.value>=Tenrs.value
after.op=before.eclairs=>before.ipp.value>=Oners.value
after.op=before.kitkat=>before.ipp.value>=Fivers.value
(before.ipp=Oners or before.ipp=Twors or before.ipp=Fivers or before.ipp=Tenrs or before.ipp=Nullrs) and
before.ipp!=Threers and before.ipp!=Fourrs and before.ipp !=Sixrs and before.ipp!=Sevenrs and before.ipp!=Eightrs and before.ipp!=Niners
(before.ip=before.b and before.ipp.value < 5) => (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum)and after.op=before.null
(before.ip=before.d and before.ipp.value < 2) => (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum)and after.op=before.null
(before.ip=before.a and before.ipp.value < 10 )=> (after.op!=before.kitkat or after.op!=before.eclairs or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.null
(before.ip=before.c and before.ipp.value >= 1) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.eclairs
(before.ip=before.c and before.ipp.value = 0) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.null
(before.ip=before.a and before.ipp.value =10) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.eclairs or after.op!=before.gum) and after.op= before.cadbury
(before.ip=before.d and before.ipp.value >= 2) => (after.op!=before.kitkat or after.op!=before.null or after.op!=before.cadbury or after.op!=before.eclairs) and after.op=before.gum
(before.ip=before.b and before.ipp.value >= 5) => (after.op!=before.eclairs or after.op!=before.null or after.op!=before.cadbury or after.op!=before.gum) and after.op=before.kitkat
}
run chocolate for exactly 2 machine, 8 button, 5 choco,9 Int,5 Coin,1 customer

Answer 1

In general, it would make sense to be more specific than "it doesn't work".
I'm assuming that what you mean by "it doesn't work" is that in the after machine you expect the quantity of the selected chocolate to be decreased by 1, but instead, it stays the same. The reason for that is your (if-then-else) or (if-then-else) or ... expression, which is logically flawed. What you probably wanted to express is to enforce at least one then branch (because you know that exactly one if condition will be satisfied), but that is not necessary to satisfy this whole disjunction.
More concretely, in
((after.op=after.cadbury)
=> (... and after.a.qty=minus[before.a.qty,1] and ...)
else (... and after.a.qty=before.a.qty and ...)
)
or
((after.op=after.kitkat)
=> (... and after.b.qty=minus[before.b.qty,1] and ...)
else (... and after.b.qty=before.b.qty and ...)
)
even if after.op is equal to after.cadbury, that does not enforce the then branch of that clause to be true, because to satisfy this whole expression, it is enough to satisfy the else branch of the next clause, which says that all quantities should stay the same.
What you want is some soft of if-then-elsif-...-else construct, e.g.,
(after.op = after.cadbury) => {
...
} else (after.op = after.kitkat) => {
...
} else {
...
}
If you do that, your machine will still not work, that is, your model will become unsatisfiable: your constraints enforce that both after and before machines share the same buttons1 and quantities are associated with buttons (the qty field is in the button sig), which means that quantities must be the same in both after and before machines. I don't really see any good reason to put qty in sig button.
[1]: by having before.cadbury=after.cadbury and ... in your chocolate predicate, and cadbury=choice.a and ... in your appended facts for sig machine