1 Sequent calculus
Your task is to implement thesequent calculus in Haskell. In the mandatory part of the project, you will omit implication → and bi-implication $ and construct proofs as explained on pages 126–127 of the textbook using the rules given there:
In the optional part, you will add rules for → and $.
Recall that proofs start with a sequent and are constructed by applying rules, working upwards, until you reach simple premises involving only“bare”predicates, not containing any connectives. The proof shows that the sequent you started with follows from those premises. If the proof ends in the empty set of simple premises, you have shown that the sequent you started with is universally valid.
Here’s an example of a sequent calculus proof, which proves that F ((¬p _ q) Λ ¬p) _p is universally valid:
Here’s another example, which proves that F ¬ ((¬a _ b) Λ (¬c _ b)) _ (¬a _ c) follows from a, b F c:
The number of times that a premise appears is immaterial, because we’re interested in the set of premises from which the conclusion holds. In this proof, the premise a, b F c happens to appear twice.
There’s often more than one proof starting from a given sequent, but all end with the same set of premises. Because each rule of the sequent calculus eliminates one connective from an antecedent or a succedent, proofs always terminate.
2 Implementing sequent calculus in Haskell
To implement the sequent calculus in Haskell, you will start with an algebraic datatype similar to the one in FP Tutorial 6 that defines propositions with variable names of type String:
type Name = String
data Prop = Var Name
| Not Prop
| Prop : || : Prop
| Prop :&&: Prop
| Prop :->: Prop
| Prop :<->: Prop
The last two cases will be required only for the optional part but they are included from the start to allow the optional part to be coded as a simple extension of the mandatory part.
Sequents are defined as follows:
data Sequent = [Prop] : |=: [Prop]
infix 0 : | =:
We use unordered lists of propositions, possibly with repetitions, to represent the sets of antecedents and succedents of sequents.
You are provided with a file Sequent .hs containing the above type definitions together with code for declaring Prop and Sequent as instances of Show and (for purposes of QuickCheck test case generation) of Arbitrary. If you want to use QuickCheck to test your code (strongly recommended! always test your code!) then you will need to comment out the following two lines for generating test cases that include → and until you get to the optional part:
, liftM2 (:->:) p2 p2
, liftM2 (:<->:) p2 p2
Don’t make any other changes to Sequent .hs.
Also provided is a skeleton file called Project .hs which imports Sequent. Your job is to define the function
prove :: Sequent -> [Sequent]
in Project .hs which, when applied to a sequent, produces the list of simple sequents from which that sequent follows according to a sequent calculus proof. You need not produce the proof itself, just the list of premises.
For the first example above, it should produce
> p = Var “p”
> q = Var “q”
> prove ([] : | =: [((Not p : || : q) :&&: Not p) : || : p])
[]
For the second example, one correct result would be
> a = Var “a”
> b = Var “b”
> c = Var “c”
> prove ([] : | =: [(Not ((Not a : || : b) :&&: (Not c : || : b))) : || : (Not a : || : c)])
[[a,b] : |=: [c]]
Because of our use of unordered lists to represent sets, another correct result would be
> prove ([] : | =: [(Not ((Not a : || : b) :&&: (Not c : || : b))) : || : (Not a : || : c)])
[[a,b] : | =: [c,c],[b,a,b] : |=: [c]]
and there are many others.
3 How to proceed
How you proceed is up to you. That’s why this is a project and not a tutorial exercise.
One way to go is to start by defining the rules of sequent calculus as Haskell functions, with for instance
orL :: Sequent -> Maybe [Sequent]
Note that the result type is Maybe [Sequent]. We use [Sequent] rather than Sequent because the _L rule has more than one premise. And we use Maybe [Sequent] rather than [Sequent] because _L isn’t applicable to all sequents, only ones having an antecedent with _ as its main connective. For the _R rule, you could define a function
orR :: Sequent -> Maybe Sequent
since it can produce at most one premise, but it’s probably better to use the same type for all rules for the sake of the next step:
orR :: Sequent -> Maybe [Sequent]
You might even want to define a type for rules:
type Rule = Sequent -> Maybe [Sequent]
orL, orR :: Rule
Hint: The main work in applying a rule to a sequent involves finding appropriate items in the antecedent list and/or the succedent list and replacing them with new items. You might find useful functions for doing this in Data.List.
Now you can use the definitions of the rules to search for a proof. Given a sequent, prove tries rules until one applies (that is, it produces Just seqs instead of Nothing). Then do that to each of the sequents in seqs . Keep going until none of the rules applies. The result is a list of simple sequents from which the original sequent follows.
The definition of prove will probably rely on helper functions that try individual rules and/or lists of rules. How you do this is up to you.
4 Optional Material: Adding implication and bi-implication
Following the Common Marking Scheme, a student with good mastery of the material is expected to get 3/4 points. This section is for demonstrating exceptional mastery of the material. It is optional and worth 1/4 points.
The definitions of the types Prop and Sequent already allow for them to contain implication → and bi-implication .
Here are the sequent calculus rules for these connectives from page 222 of the textbook:
Here is a proof that a → b F ¬b → ¬a is universally valid, using these rules and the ones given earlier:
For the optional part, extend your function
prove :: Sequent -> [Sequent]
to dosequent calculus proofs of sequents involving propositions that may contain implication and/or bi-implication. Applying prove to asequent that doesn’t contain implication orbi-implication should produce the same result as before. If you do the optional part, you don’t have to submit two versions of prove; the extended version, that works for implications and bi-implications as well as the other connectives, will suffice.
For the example above, it should produce
> a = Var “a”
> b = Var “b”
> prove (a :->: b : |=: Not b :->: Not a)
[]
If you use QuickCheck for testing your code, remember to restore the two lines of the declaration in Sequent .hs that Prop is an instance of Arbitrary so that it will also generate test cases including :->: and :<->:.
6 Marking
The programming project will be marked by your tutor, and is worth 20% of the mark for Inf1A. Your mark on the scale 0–4 will be based mainly on the following tests. (Your tutor will also check your code to make sure, among other things, that it is not written to pass only these tests.) The tests have been set up in the automarker. They will be run every time you submit Project .hs, and you will be able to check the results. You can submit as often as you like, but only the last submission before the deadline will be taken into account in the marking.
The sequents in the tests below have been typeset to make them a little easier to read, but of course the actual tests use Haskell syntax. Because of our use of unordered lists to represent tests, lists of assumptions that are equivalent to those shown as results are also correct.
To get 1 point, it’s enough that any one of the following one-step proofs is correct. Non-working code that shows some sensible work may also be awarded 1 point.
> prove ( F a _ b ) — _R
[ F a, b ]
> prove ( a _ b F ) — _L
[ a F , b F ]
> prove ( F a Λ b ) — ΛR
[ F a , F b ]
> prove ( a Λ b F ) — ΛL
[ a, b F ]
> prove ( F ¬a ) — ¬R
[ a F ]
> prove ( ¬a F ) — ¬L
[ F a ]
> prove ( a,b F b,c ) — I
[ ]
To get 2 points, the following proof, which uses only the non-branching rules, must also be correct.
> prove ( F (¬ (e Λ (f Λ ¬¬c))) _ (¬a _ c) ) [ ]
To get 3 points, most or all of the following proofs must also be correct.
> prove ( F ((¬a _ b) Λ ¬a) _ a ) [ ]
> prove ( F ¬ ((¬a _ b) Λ (¬c _ b)) _ ¬a _ c ) [ a, b F c ]
> prove ( ¬c _ (f _ b), (a _ d) Λ (b _ b) F ¬ (d _ b), ¬c _ (f Λ e), (f Λ b) _ (c Λ e), ¬ (a _ c), ¬e Λ (c _ e), e ) [ ]
> prove ( (b _ f) _ (d _ c), ¬ (d _ a), ¬ (d Λ f), ¬ (a Λ f), (a _ e) Λ (b _ c), c F )
[ b,c, e F a,d , b,c, e F a,d,f , b,c,e, f F a,d , c, e F a,d , c, e F a,d,f , c,e, f F a, d ]
> prove ( b,(d _ b) _ ¬c, ¬ (d Λ f) F (b _ d) Λ (c Λ e), (a _ c) _ ¬f, ¬ (f Λ f), (e Λ d) Λ (a _ d),c, (b Λ b) Λ (e _ b) ) [ ]
> prove ( (f _ a) Λ ¬f,(a _ a) Λ (c _ b), ¬a _ ¬c, ¬d _ (a _ d)
F (f Λ d) _ (d _ d), (c _ e) _ (d _ e), (b _ a) Λ (f _ e) )
[ a, b F c,d,e, f ]
> prove ( (a _ b) _ ¬d,(e _ b) Λ ¬f F a, (f Λ f) _ (f Λ f), ¬f Λ (e _ c), (f Λ b) Λ (e _ a) ) [ b F a,e,c,f , b F a,c,d,e, f ]
> prove ( (a _ f) _ (c _ d), (b Λ f) _ (f _ a), (f _ f) Λ ¬f, ¬a _ (c _ e) F ¬e Λ ¬c, (b Λ c) _ ¬a, ¬ (e _ d) ) [ ]
> prove ( F (d _ b) _ (e Λ a), ¬f Λ (d Λ b), (c Λ b) Λ ¬e, (c _ c) _ (d _ d),
¬e Λ (a Λ e), (a _ e) _ (c _ d), (d _ d) Λ ¬e, (c Λ d) Λ ¬a, ¬ (e _ c) ) [ ]
To get 4 points, most or all of the following proofs involving implication and/or bi-implication must also be correct.
> prove ( a → b F ¬b → ¬a ) [ ]
> prove ( (c _ d) → (d Λ f), (a a) (b _ a) F (c b) (b → a), ¬ (b _ d), ¬ (b Λ d), (d f) → (c _ e) ) [ b,a,f,d F c, e ]
> prove ( ¬ (f f), ¬f (d e), (d → f) (a _ b) F (d Λ c) → (c d), ¬ (c e) ) [ ]
> prove ( (e → b) → (a → f), (b _ a) _ ¬f,(d b) (f Λ e) F (d d) _ (e c), ¬ (d → f) ) [ ]
> prove ( (f _ b) Λ (b → d), (e _ d) _ ¬e, (f → e) → ¬f,(e _ d) Λ (d c), ¬ (c b), ¬f Λ (a Λ a),
(f c) → (c _ f) F (e Λ f) _ (f Λ a), ¬f Λ ¬d,(f _ c) (f b) ) [ ]
> prove ( (e Λ f) (a Λ e), (b _ d) Λ (f c), ¬ (d f), (d Λ e) _ (c → f), (e f) (b _ d) F (e _ d) Λ ¬e ) [ f,e,a,b, c F d ]
The challenge part isn’t worth any points, but the following tests are included in the automarker. Correct for these means getting the same assumptions with a proof that is no longer than the number of rules shown, which is achieved by the sample solution.
> proveCount ( (a _ f) _ (c _ d), (b Λ f) _ (f _ a), (f _ f) Λ ¬f, ¬a _ (c _ e) F ¬e Λ ¬c, (b Λ c) _ ¬a, ¬ (e _ d) ) ([ ], 168)
> proveCount ( ¬c _ (f _ b), (a _ d) Λ (b _ b) F ¬ (d _ b), ¬c _ (f Λ e), (f Λ b) _ (c Λ e), ¬ (a _ c), ¬e Λ (c _ e), e ) ([ ], 39)
> proveCount ( b,(d _ b) _ ¬c, ¬ (d Λ f) F (b _ d) Λ (c Λ e), (a _ c) _ ¬f, ¬ (f Λ f), (e Λ d) Λ (a _ d),c, (b Λ b) Λ (e _ b) ([ ], 58)
> proveCount ( (c _ d) → (d Λ f), (a a) (b _ a) F (c b) (b → a), ¬ (b _ d), ¬ (b Λ d), (d f) → (c _ e) ) ([ a,b,f,d F c, e ], 144)
> proveCount ( ¬ (f f), ¬f (d e), (d → f) (a _ b) F (d Λ c) → (c d), ¬ (c e) ) ([ ], 105)
> proveCount ( (e → b) → (a → f), (b _ a) _ ¬f,(d b) (f Λ e) F (d d) _ (e c), ¬ (d → f) ) ([ ], 306)
> proveCount ( (f _ b) Λ (b → d), (e _ d) _ ¬e, (f → e) → ¬f,(e _ d) Λ (d c), ¬ (c b), ¬f Λ (a Λ a),
(f c) → (c _ f) F (e Λ f) _ (f Λ a), ¬f Λ ¬d,(f _ c) (f b) )
([ ], 229)
> proveCount ( (e Λ f) (a Λ e), (b _ d) Λ (f c), ¬ (d f), (d Λ e) _ (c → f), (e f) (b _ d) F (e _ d) Λ ¬e ) ([ a,b,c,e, f F d ], 491)
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