An amazing feature of Imandra is the ability to reason about functional inputs and generate functional instances and counterexamples. For example we can introduce a variant type and a second-order function that evaluates its argument on each variant and computes the sum of the results.

In :
type t = A | B | C
let sum = (f) => f(A) + f(B) + f(C)

Out:
type t = A | B | C
val sum : (t -> Z.t) -> Z.t = <fun>


Now, let's ask Imandra to come up with an instance of a function f that returns positive values and that produces 42 when calling sum(f):

In :
instance( (f) => f(A) > 0 && f(B) > 0 && sum(f) == 42 )

Out:
- : (t -> int) -> bool = <fun>
module CX : sig val f : t -> Z.t end

Instance (after 0 steps, 0.024s):
let f = function
| C -> 40
| _ -> 1


Instance
proof attempt
ground_instances0
definitions0
inductions0
search_time
0.024s
details
Expand
smt_stats
 num checks 1 arith assert lower 3 arith pivots 1 rlimit count 235 mk clause 3 datatype occurs check 3 mk bool var 11 arith assert upper 1 propagations 2 datatype accessor ax 3 num allocs 2.34996e+06 final checks 1 added eqs 1 del clause 3 arith eq adapter 1 memory 5.88 max memory 6.28
Expand
• start[0.024s]
sko_f_0 A > 0 && sko_f_0 B > 0 && (sko_f_0 A + sko_f_0 B) + sko_f_0 C = 42
• #### simplify

 into (not (sko_f_0 A <= 0) && not (sko_f_0 B <= 0)) && (sko_f_0 A + sko_f_0 B) + sko_f_0 C = 42 expansions [] rewrite_steps forward_chaining
• Sat (Some let f = function | C -> 40 | _ -> 1 )

A natural application of this ability is to apply it to formal reasoning about random variables and probability distributions over them. In the following, we'll consider an application of this kind of reasoning to analyse the famous Monty Hall problem.

# Reasoning about Monty Hall probabilities¶

Let’s ask Imandra if 1/3 of a 1,000,000$is the best outcome for any strategy the player can employ. To do this we’ll use Imandra verify directive to verify that, for any valid strategy, the scenario with equally probable prize placement gives an expected reward less or equal than 1000000/3. In : verify ( (strategy) => Real.({ valid_strategy(strategy) ==> ( expected_reward ( make_scenario_pmf(pDoorsEqual,strategy) ) <= 1000000.0 / 3.0 ) }))  Out: - : (door -> choice -> real) -> bool = <fun> module CX : sig val strategy : door -> choice -> Q.t end  Counterexample (after 0 steps, 0.029s): let strategy x_0 x_1 = match (x_0, x_1) with | (DoorA, Swap) -> (Real.mk_of_string "3/1000000") | (DoorB, Stay) -> (Real.mk_of_string "999997/1000000") | _ -> 0.0  Refuted proof attempt ground_instances0 definitions0 inductions0 search_time 0.029s details Expand smt_stats  num checks 1 arith assert lower 8 arith pivots 2 rlimit count 1557 mk clause 3 datatype occurs check 5 mk bool var 16 arith assert upper 1 arith add rows 2 propagations 2 datatype accessor ax 5 num allocs 3.53524e+06 final checks 1 added eqs 1 del clause 3 arith eq adapter 1 memory 8.09 max memory 8.57 Expand • start[0.029s] sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0 && sko_strategy_0 DoorB Stay >=. 0 && sko_strategy_0 DoorB Swap >=. 0 && sko_strategy_0 DoorC Stay >=. 0 && sko_strategy_0 DoorC Swap >=. 0 && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1 ==> ((((((1 /. 3 *. sko_strategy_0 DoorA Stay) *. (if DoorA = DoorA then if Stay = Stay then 1000000 else 0 else if Stay = Swap then 1000000 else 0) +. (1 /. 3 *. sko_strategy_0 DoorA Stay) *. (if DoorB = DoorA then if Stay = Stay then 1000000 else 0 else if … then 1000000 else 0)) +. …) +. …) +. …) +. …) <=. … • #### simplify  into not ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) || (((((1000000/3 *. sko_strategy_0 DoorA Stay +. 1000000/3 *. sko_strategy_0 DoorB Stay) +. 1000000/3 *. sko_strategy_0 DoorC Stay) +. 2000000/3 *. sko_strategy_0 DoorA Swap) +. 2000000/3 *. sko_strategy_0 DoorB Swap) +. 2000000/3 *. sko_strategy_0 DoorC Swap) <=. 1000000/3 expansions [] rewrite_steps forward_chaining • Sat (Some let strategy x_0 x_1 = match (x_0, x_1) with | (DoorA, Swap) -> (Real.mk_of_string "3/1000000") | (DoorB, Stay) -> (Real.mk_of_string "999997/1000000") | _ -> 0.0 ) Imandra refuted our statement and provided us with a counterexample strategy. Imandra suggests that the expected reward will be greater than 1000000/3 if the player chooses DoorA and then Swaps his choice in 0.0003% of the cases. The counterexample strategy is defined in the CX module, so let's see what is the expected reward with this strategy: In : expected_reward( make_scenario_pmf(pDoorsEqual, CX.strategy) )  Out: - : real = 1000003/3  The counterexample strategy that Imandra produced increased our expected winnings by 1$. This counterexample strategy employs swapping instead of staying. What would be our expected reward if we always swap?

In :
let random_then_swap_strategy = (door : door,choice) =>
switch(door,choice){ | (_,Swap) => Real.(1.0 / 3.0) | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsEqual,random_then_swap_strategy)
)

Out:
val random_then_swap_strategy : door -> choice -> real = <fun>
- : real = 2000000/3


By always swapping we’ve increased our chances of winning up to 2/3. Is this the best possible expected reward one might get in this game?

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsEqual,strategy)
) <= 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances0
definitions0
inductions0
search_time
0.021s
details
Expand
smt_stats
 num checks 1 arith assert lower 8 arith pivots 2 rlimit count 1538 mk clause 3 mk bool var 16 arith assert upper 1 arith add rows 2 propagations 2 conflicts 1 datatype accessor ax 5 arith conflicts 1 num allocs 9.7614e+06 added eqs 1 arith eq adapter 1 memory 11.34 max memory 11.73
Expand
• start[0.021s]
sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1
==> ((((((1 /. 3 *. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA then if Stay = Stay then 1000000 else 0
else if Stay = Swap then 1000000 else 0)
+. (1 /. 3 *. sko_strategy_0 DoorA Stay)
*. (if DoorB = DoorA then if Stay = Stay then 1000000 else 0
else if … then 1000000 else 0))
+. …)
+. …)
+. …)
+. …)
<=. …
• #### simplify

 into not ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) || (((((1000000/3 *. sko_strategy_0 DoorA Stay +. 1000000/3 *. sko_strategy_0 DoorB Stay) +. 1000000/3 *. sko_strategy_0 DoorC Stay) +. 2000000/3 *. sko_strategy_0 DoorA Swap) +. 2000000/3 *. sko_strategy_0 DoorB Swap) +. 2000000/3 *. sko_strategy_0 DoorC Swap) <=. 2000000/3 expansions [] rewrite_steps forward_chaining
• #### unsat

(let ((a!1 (+ (sko_strategy_0_1257 DoorA_24 Stay_28)
(sko_strategy_0_1257 DoorA_24 Swa…

We’ve proven that the best possible expected reward in the Monty Hall game is 2/3 of a \$1,000,000 in an (arguably, rather counterintuitive) result that one has to always swap his first choice to double his chances of winning.

Suppose now that we don’t know the winning strategy, but we do know that it is possible to win in 2/3 of the cases. Can we ask Imandra to synthesise the optimal strategy? For this purpose we can use the instance directive:

In :
instance ( (strategy) => Real.({
valid_strategy(strategy) &&
( expected_reward (
make_scenario_pmf(pDoorsEqual,strategy)
) == 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> Q.t end

Instance (after 0 steps, 0.039s):
let strategy x_0 x_1 =
match (x_0, x_1) with
| (DoorA, Swap) -> 1.0
| _ -> 0.0


Instance
proof attempt
ground_instances0
definitions0
inductions0
search_time
0.039s
details
Expand
smt_stats
 num checks 1 arith assert lower 8 arith pivots 2 rlimit count 1547 mk clause 6 datatype occurs check 5 mk bool var 18 arith assert upper 2 arith add rows 2 propagations 4 datatype accessor ax 5 num allocs 3.14604e+07 final checks 1 added eqs 2 del clause 6 arith eq adapter 2 memory 14.45 max memory 14.94
Expand
• start[0.039s]
(sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1)
&& (((((1 /. 3 *. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA then if Stay = Stay then 1000000 else 0
else if Stay = Swap then 1000000 else 0)
+. (1 /. 3 *. sko_strategy_0 DoorA Stay)
*. (if DoorB = DoorA then if Stay = Stay then 1000000 else 0
else if … then 1000000 else 0))
+. …)
+. …)
+. …)
+. … = …
• #### simplify

 into ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) && ((((1000000/3 *. sko_strategy_0 DoorA Stay +. 1000000/3 *. sko_strategy_0 DoorB Stay) +. 1000000/3 *. sko_strategy_0 DoorC Stay) +. 2000000/3 *. sko_strategy_0 DoorA Swap) +. 2000000/3 *. sko_strategy_0 DoorB Swap) +. 2000000/3 *. sko_strategy_0 DoorC Swap = 2000000/3 expansions [] rewrite_steps forward_chaining
• Sat (Some let strategy x_0 x_1 = match (x_0, x_1) with | (DoorA, Swap) -> 1.0 | _ -> 0.0 )

We've got an example of an optimal strategy - this one doesn't randomize over doors, but suggests to always choose the DoorA and then swap.

# Biased Monty Hall¶

Now let's apply what we've learned about the "standard" Monty Hall paradox to a more complicated game setup - suppose that we somehow know that the host is biased towards the DoorA -- the probability that the host will hide the prize behind the first door is 3 times higher than the individual probabilities for two other doors. Should one still always swap? Or, maybe, it makes more sense to always choose the more likely DoorA and stay?

We encode our biased host in the biased PMF over doors (also doing a sanity check that it is a valid probability distribution):

In :
let pDoorsBiased = (door : door) => Real.(
switch (door){
| DoorA => 3.0 / 5.0
| DoorB | DoorC => 1.0 / 5.0
}
);

valid_door_pmf(pDoorsBiased)

Out:
val pDoorsBiased : door -> real = <fun>
- : bool = true


In the previous section, we've proven that the maximal possible expected reward in the "standard" Monty Hall problem is 2/3 of a million. Again, we ask Imandra whether we can improve on this in the biased case:

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsBiased,strategy)
) <= 2000000.0 / 3.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> Q.t end

Counterexample (after 0 steps, 0.023s):
let strategy x_0 x_1 =
match (x_0, x_1) with
| (DoorA, Stay) -> (Real.mk_of_string "399997/600000")
| (DoorC, Swap) -> (Real.mk_of_string "200003/600000")
| _ -> 0.0


Refuted
proof attempt
ground_instances0
definitions0
inductions0
search_time
0.023s
details
Expand
smt_stats
 num checks 1 arith assert lower 8 arith pivots 3 rlimit count 1857 mk clause 3 datatype occurs check 5 mk bool var 16 arith assert upper 1 arith add rows 3 propagations 2 datatype accessor ax 5 num allocs 2.0431e+07 final checks 1 added eqs 1 del clause 3 arith eq adapter 1 memory 14.87 max memory 15.35
Expand
• start[0.023s]
sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1
==> (((((((if DoorA = DoorA then 3 /. 5 else 1 /. 5)
*. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA then if Stay = Stay then 1000000 else 0
else if Stay = Swap then 1000000 else 0)
+. ((if DoorB = DoorA then 3 /. 5 else 1 /. 5)
*. sko_strategy_0 DoorA Stay)
*. (if … then … else …))
+. …)
+. …)
+. …)
+. …)
<=. …
• #### simplify

 into not ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) || (((((600000 *. sko_strategy_0 DoorA Stay +. 200000 *. sko_strategy_0 DoorB Stay) +. 200000 *. sko_strategy_0 DoorC Stay) +. 400000 *. sko_strategy_0 DoorA Swap) +. 800000 *. sko_strategy_0 DoorB Swap) +. 800000 *. sko_strategy_0 DoorC Swap) <=. 2000000/3 expansions [] rewrite_steps forward_chaining
• Sat (Some let strategy x_0 x_1 = match (x_0, x_1) with | (DoorA, Stay) -> (Real.mk_of_string "399997/600000") | (DoorC, Swap) -> (Real.mk_of_string "200003/600000") | _ -> 0.0 )

Imandra suggests that we can do better than 2/3 if we use a mixed strategy of either choosing the DoorA and staying or choosing the DoorC and swapping. Let's test both of these pure strategies and see what would be the expected rewards for each:

In :
let aStayStrategy = (door,choice) =>
switch(door,choice){ | (DoorA,Stay) => 1.0 | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsBiased,aStayStrategy)
);
let cSwapStrategy = (door,choice) =>
switch(door,choice){ | (DoorC,Swap) => 1.0 | _ => 0.0 };
expected_reward(
make_scenario_pmf(pDoorsBiased,cSwapStrategy)
);

Out:
val aStayStrategy : door -> choice -> Q.t = <fun>
- : real = 600000
val cSwapStrategy : door -> choice -> Q.t = <fun>
- : real = 800000


Choosing the DoorC and then swapping increases the chances of winning up to 80%! Again, is this the best outcome one can get?

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward (
make_scenario_pmf(pDoorsBiased,strategy)
) <= 800000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances0
definitions0
inductions0
search_time
0.043s
details
Expand
smt_stats
 num checks 1 arith assert lower 8 arith pivots 3 rlimit count 1836 mk clause 3 mk bool var 16 arith assert upper 1 arith add rows 3 propagations 2 conflicts 1 datatype accessor ax 5 arith conflicts 1 num allocs 3.45092e+07 added eqs 1 arith eq adapter 1 memory 18.31 max memory 18.8
Expand
• start[0.043s]
sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1
==> (((((((if DoorA = DoorA then 3 /. 5 else 1 /. 5)
*. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA then if Stay = Stay then 1000000 else 0
else if Stay = Swap then 1000000 else 0)
+. ((if DoorB = DoorA then 3 /. 5 else 1 /. 5)
*. sko_strategy_0 DoorA Stay)
*. (if … then … else …))
+. …)
+. …)
+. …)
+. …)
<=. 800000
• #### simplify

 into not ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) || (((((600000 *. sko_strategy_0 DoorA Stay +. 200000 *. sko_strategy_0 DoorB Stay) +. 200000 *. sko_strategy_0 DoorC Stay) +. 400000 *. sko_strategy_0 DoorA Swap) +. 800000 *. sko_strategy_0 DoorB Swap) +. 800000 *. sko_strategy_0 DoorC Swap) <=. 800000 expansions [] rewrite_steps forward_chaining
• #### unsat

(let ((a!1 (+ (* 600000.0 (sko_strategy_0_1331 DoorA_24 Stay_28))
(* 200000.0 (sko_str…

And we've proven that the DoorC-Swap is the best strategy the player can employ in this setup. Intuitively the reason for that is that it is better to choose the least likely door, forcing the host to reveal more information about the prize location in the more likely doors.

# Even more complex example¶

Using this approach one can go further and analyse even more complicated Monty-Hall-like games (for example, the prize value might depend on the door and/or player actions).

In :
let pDoorsComplex = (door : door) => Real.(
switch (door){
| DoorA => 5.0 / 10.0
| DoorB => 3.0 / 10.0
| DoorC => 2.0 / 10.0
}
);

let reward_complex = (scenario) => {
let prize = switch(scenario.prize) {
| DoorA => 1000000.
| DoorB => 2000000.
| DoorC => 4000000.
};
if( scenario.prize == scenario.first_guess ) {
if(scenario.choice == Stay) prize else 0.0;
} else {
if(scenario.choice == Swap) prize else 0.0;
}
};

let expected_reward_complex : (scenario => real) => real =
(scenario_pmf) => Real.({
let pr = (s) => scenario_pmf(s) * reward_complex(s);
let avg = (choice) => {
let avg = (first_guess) => {
pr { prize: DoorA, first_guess, choice} +
pr { prize: DoorB, first_guess, choice} +
pr { prize: DoorC, first_guess, choice}
};
avg(DoorA) + avg(DoorB) + avg(DoorC)
};
avg(Stay) + avg(Swap)
})

Out:
val pDoorsComplex : door -> real = <fun>
val reward_complex : scenario -> Q.t = <fun>
val expected_reward_complex : (scenario -> real) -> real = <fun>

In :
verify ( (strategy) => Real.({
valid_strategy(strategy) ==>
( expected_reward_complex (
make_scenario_pmf(pDoorsComplex,strategy)
) <= 1400000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>

Proved
proof
ground_instances0
definitions0
inductions0
search_time
0.020s
details
Expand
smt_stats
 num checks 1 arith assert lower 8 arith pivots 2 rlimit count 2238 mk clause 3 mk bool var 16 arith assert upper 1 arith add rows 2 propagations 2 conflicts 1 datatype accessor ax 5 arith conflicts 1 num allocs 5.3011e+07 added eqs 1 arith eq adapter 1 memory 22.04 max memory 22.52
Expand
• start[0.020s]
sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1
==> (((((((if DoorA = DoorA then 5 /. 10
else if DoorA = DoorB then 3 /. 10 else 2 /. 10)
*. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA
then
if Stay = Stay
then
if DoorA = DoorA then 1000000
else if DoorA = DoorB then 2000000 else 4000000
else 0
else
if Stay = Swap then if DoorA = DoorA then 1000000 else …
else 0)
+. …)
+. …)
+. …)
+. …)
+. …)
<=. 1400000
• #### simplify

 into not ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) || (((((500000 *. sko_strategy_0 DoorA Stay +. 600000 *. sko_strategy_0 DoorB Stay) +. 800000 *. sko_strategy_0 DoorC Stay) +. 1400000 *. sko_strategy_0 DoorA Swap) +. 1300000 *. sko_strategy_0 DoorB Swap) +. 1100000 *. sko_strategy_0 DoorC Swap) <=. 1400000 expansions [] rewrite_steps forward_chaining
• #### unsat

(let ((a!1 (+ (sko_strategy_0_1379 DoorA_24 Stay_28)
(sko_strategy_0_1379 DoorA_24 Swa…
In :
instance ( (strategy) => Real.({
valid_strategy(strategy) &&
( expected_reward_complex (
make_scenario_pmf(pDoorsComplex,strategy)
) == 1400000.0 )
}))

Out:
- : (door -> choice -> real) -> bool = <fun>
module CX : sig val strategy : door -> choice -> Q.t end

Instance (after 0 steps, 0.029s):
let strategy x_0 x_1 =
match (x_0, x_1) with
| (DoorA, Swap) -> 1.0
| _ -> 0.0


Instance
proof attempt
ground_instances0
definitions0
inductions0
search_time
0.029s
details
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smt_stats
 num checks 1 arith assert lower 8 arith pivots 2 rlimit count 2247 mk clause 6 datatype occurs check 5 mk bool var 18 arith assert upper 2 arith add rows 2 propagations 4 datatype accessor ax 5 num allocs 1.19767e+08 final checks 1 added eqs 2 del clause 6 arith eq adapter 2 memory 27.8 max memory 28.29
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• start[0.029s]
(sko_strategy_0 DoorA Stay >=. 0
&& sko_strategy_0 DoorA Swap >=. 0
&& sko_strategy_0 DoorB Stay >=. 0
&& sko_strategy_0 DoorB Swap >=. 0
&& sko_strategy_0 DoorC Stay >=. 0
&& sko_strategy_0 DoorC Swap >=. 0
&& ((((sko_strategy_0 DoorA Stay
+. sko_strategy_0 DoorA Swap)
+. sko_strategy_0 DoorB Stay)
+. sko_strategy_0 DoorB Swap)
+. sko_strategy_0 DoorC Stay)
+. sko_strategy_0 DoorC Swap = 1)
&& ((((((if DoorA = DoorA then 5 /. 10
else if DoorA = DoorB then 3 /. 10 else 2 /. 10)
*. sko_strategy_0 DoorA Stay)
*. (if DoorA = DoorA
then
if Stay = Stay
then
if DoorA = DoorA then 1000000
else if DoorA = DoorB then 2000000 else 4000000
else 0
else
if Stay = Swap then if DoorA = DoorA then 1000000 else …
else 0)
+. …)
+. …)
+. …)
+. …)
+. … = 1400000
• #### simplify

 into ((((((sko_strategy_0 DoorA Stay >=. 0 && sko_strategy_0 DoorA Swap >=. 0) && sko_strategy_0 DoorB Stay >=. 0) && sko_strategy_0 DoorB Swap >=. 0) && sko_strategy_0 DoorC Stay >=. 0) && sko_strategy_0 DoorC Swap >=. 0) && ((((sko_strategy_0 DoorA Stay +. sko_strategy_0 DoorA Swap) +. sko_strategy_0 DoorB Stay) +. sko_strategy_0 DoorB Swap) +. sko_strategy_0 DoorC Stay) +. sko_strategy_0 DoorC Swap = 1) && ((((500000 *. sko_strategy_0 DoorA Stay +. 600000 *. sko_strategy_0 DoorB Stay) +. 800000 *. sko_strategy_0 DoorC Stay) +. 1400000 *. sko_strategy_0 DoorA Swap) +. 1300000 *. sko_strategy_0 DoorB Swap) +. 1100000 *. sko_strategy_0 DoorC Swap = 1400000 expansions [] rewrite_steps forward_chaining
• Sat (Some let strategy x_0 x_1 = match (x_0, x_1) with | (DoorA, Swap) -> 1.0 | _ -> 0.0 )

More generally, this illustrates how we can use Imandra’s ability to reason about functional values and real arithmetic to formally prove and refute statements about probability distributions. These abilities allow us to us to formally analyse many kinds of scenarios requiring the ability to reason about random variables.