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# Proving Theorems in the ProB REPL

For this example we try and use the REPL (Read-Eval-Print-Loop) of ProB to prove theorems. The REPL can either be started using probcli's command -repl or by starting the Eval Console in ProB Tcl/Tk.

See the beginning of Sudoku Solved in the ProB REPL for more details about how to start the REPL.

The general principle for proving a sequent Hyp |- Goal is to state the hypotheses Hyp and negate the goal. If ProB finds a solution, a counter example to the sequent exists and it cannot be proven. If ProB finds no solution, then under certain circumstances the sequent is proven, as no counter example exists. (This mechanism is used by the ProB Disprover in Rodin.)

Let us prove that from x:1..100 it follows that the equation x*x+2*x+2 = 1 has no solution:

```>>> x:1..100 & not(x*x+2*x+2/=1)
Existentially Quantified Predicate over x is FALSE
```

In this case, ProB has not generated an "enumeration warning", i.e., all cases where considered: we have a proof. In general when all variables are restricted to a finite set of values in the hypotheses, ProB can be used as a prover.

Let us now lift the finiteness restriction on x and prove the theorem for all integer values:

```>>> x:INTEGER & not(x*x+2*x+2/=1)
### Warning: enumerating x : INTEGER : inf: -1 ---> -1: -1
Existentially Quantified Predicate over x is TRUE
Solution:
x = -1
```

Here ProB has found a counter example and the sequent is not valid. As you can see, an enumeration warning occurred; meaning that not all of the infinitely many values for x will be tried. But this warning is not relevant here as a solution was found.

Let us try and restrict x to positive values; the number of values are infinite (NATURAL is the set of mathematical natural numbers, not restricted by MAXINT):

```>>> x:NATURAL & not(x*x+2*x+2/=1)
Existentially Quantified Predicate over x is FALSE
```

Here ProB has found no solution and generated no enumeration warning: the sequent is proven.

Now let us try and prove that x>=y follows from x>y:

```>>> x>y & not(x>=y)
% Timeout when posting constraint:
% kernel_objects:(_15735#>0)
### Warning: enumerating x : INTEGER : inf:sup ---> -1:3
% Timeout when posting constraint:
% kernel_objects:(_15735#<0)
Existentially Quantified Predicate over x,y is UNKNOWN [FALSE with ** ENUMERATION WARNING **]
```

As you can see, here the solver was not able to detect the inconsistency and prove the sequent over the unbounded integers x and y. There are two things one can try. First, one can enable some additional constraint propagation rules (implemented using the Constraint Handling Rules library of Prolog) in the standard ProB kernel by setting the CHR preference to TRUE:

```>>> -p CHR TRUE
Executing probcli command: [set_pref(CHR,TRUE)]
>>> x>y & not(x>=y)
Existentially Quantified Predicate over x,y is FALSE
```

This has enabled us to prove the sequent. (The ProB Disprover automatically sets this preference for you.) Another solution is again to use the Z3 backend:

```>>> :z3 x>y & not(x>=y)
PREDICATE is FALSE
```

### User-Defined Sets

Suppose we want to prove a property not over a specific set such as the integers but over an arbitrary set. If you want to use additional basic types in the ProB REPL you need to define them in a B machine, they cannot (yet) be added in the REPL. For example, let us create a file DefSets.mch:

```MACHINE DefSets
SETS D;E = {e1,e2,e3,e4,e5} // we just define two sets to be used in the REPL, D is deferred, E is enumerated
END
```

Then you can load the file and start the REPL:

```rlwrap probcli DefSets.mch -repl
```

Let us try and prove that A={a,b} & B={c,d} implies A /\ B = {}. We can state this as follows. Note that in B we need to type A and B. Here we first use the enumerated set E for that purpose.

```>>> A <: E & B<: E & A={a,b} & B={c,d} & not(  A /\ B = {} )
Existentially Quantified Predicate over A,B,a,b,c,d is TRUE
Solution:
A = {e1} &
B = {e1} &
a = e1 &
b = e1 &
c = e1 &
d = e1
```

As you can see, the goal A /\ B = {} cannot be proven, a counter-example is provided. If we add hypotheses that a,b,c,d are all pair-wise distinct we can prove the assertion:

```>>> A <: E & B<: E & A={a,b} & B={c,d} & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d & not(  A /\ B = {} )
Existentially Quantified Predicate over A,B,a,b,c,d is FALSE
```

This has used ProB’s Prolog-CLPFD solver. You can use ProB's translation to Z3 (see Using ProB with Z3) by prefixing the formula with z3:

```>>> :z3 A <: E & B<: E & A={a,b} & B={c,d} & not(  A /\ B = {} ) & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d
PREDICATE is FALSE
```

You can also translate the query to SAT via the Kodkod library (see Using ProB with KODKOD) using:

```>>> :kodkod A <: E & B<: E & A={a,b} & B={c,d} & not(  A /\ B = {} ) & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d
kodkod ok:   A = {a,b} & B = {c,d} & not(A /\ B = {}) & a /= b ...  ints: none, intatoms: none
Kodkod module started up successfully (SAT solver SAT4J with timeout of 1500 ms).
Times for computing solutions: []
Existentially Quantified Predicate over A,B,a,b,c,d is FALSE
```

Note that ProB’s Prolog-CLPFD solver and Kodkod cannot provide proofs when you use deferred sets whose cardinality is not specified and finite. For example, let us use the deferred set D instead of E to type A and B:

```>>> A <: D & B<: D & A={a,b} & B={c,d} & not(  A /\ B = {} ) & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d
Existentially Quantified Predicate over A,B,a,b,c,d is FALSE
>>> D
Expression Value =
{D1,D2}
```

As you can see, ProB has set the size of the set D to two (see Deferrred_Sets). You can see this also by asking ProB to find four different Values:

```>>> A <: D & B<: D & A={a,b} & B={c,d} & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d
Existentially Quantified Predicate over A,B,a,b,c,d is FALSE
```

So, in the presence of deferred sets, ProB's default solver and the Kodkod-based solver cannot be used to prove theorems; they can only be used to find counter-examples. The Z3 backend however does not fix the size of D:

```>>> :z3 A <: D & B<: D & A={a,b} & B={c,d} & a/=b & a/=c & a/=d & b/=c & b/=d & c/=d
PREDICATE is TRUE
Solution:
A = {D1,D2}
B = {}
a = D1
b = D2
c = D3
d = D4
```