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106 lines
4.6 KiB
106 lines
4.6 KiB
> Show me an implementation of the factorial algorithm in Coq




Certainly! Here is an implementation of the factorial algorithm in Coq:




python




Fixpoint factorial (n : nat) : nat :=


match n with


 O => 1


 S n' => n * factorial n'


end.




In this implementation, factorial is a recursive function that takes a natural number n as an argument and returns the factorial of n. The base case of the recursion is when n is 0, in which case the function returns 1. Otherwise, the function recursively calls itself with n', which is n decremented by 1, and multiplies the result by n.




Here's an example of how to use this function in Coq:




java




Eval compute in (factorial 5).




This will evaluate to 120, which is the factorial of 5.




> Now, show me an implementation of the factorial algorithm in Coq using an inductive relation instead of a recursive function.




Sure, here's an implementation of the factorial algorithm using an inductive relation:




rust




Inductive factorial : nat > nat > Prop :=


 fact_base : factorial O 1


 fact_step : forall n m, factorial n m > factorial (S n) ((S n) * m).




In this implementation, factorial is an inductive relation that takes two arguments: a natural number n and its factorial m. The fact_base constructor defines the base case for the relation when n is 0, and m is 1. The fact_step constructor defines the recursive step for the relation when n is incremented by 1, and m is updated by multiplying m by (n+1).




Here's an example of how to use this relation in Coq:




cpp




Example factorial_5 : factorial 5 120.


Proof.


apply (fact_step 4 24).


apply (fact_step 3 6).


apply (fact_step 2 2).


apply (fact_step 1 1).


apply fact_base.


Qed.




This shows that the relation holds for n = 5 and m = 120. Each fact_step constructor represents a step in the calculation of the factorial, and the fact_base constructor represents the base case when n = 0 and m = 1.




> Now, show me a proof that the inductive relation is equivalent to the recursive function definition.




Certainly! Here's a proof that the inductive relation and recursive function definitions of factorial are equivalent:




java




Lemma factorial_equiv : forall n, factorial n (factorial_rec n).


Proof.


induction n.


 apply fact_base.


 simpl.


apply (fact_step n (factorial_rec n)).


apply IHn.


Qed.




Lemma factorial_rec_equiv : forall n, factorial_rec n = factorial n.


Proof.


intros n.


unfold factorial_rec.


induction n.


 reflexivity.


 simpl.


rewrite IHn.


apply (fact_step n (factorial n)).


apply factorial_equiv.


Qed.




The first lemma factorial_equiv shows that the inductive relation holds for every n with the corresponding value of the recursive function, i.e., factorial n (factorial_rec n). This is proved by induction on n, where the base case is fact_base and the recursive case is fact_step using the induction hypothesis.




The second lemma factorial_rec_equiv shows that the recursive function definition of factorial is equivalent to the inductive relation definition. This is proved by induction on n, where the base case is trivial and the recursive case uses the induction hypothesis to rewrite the factorial of n in terms of the factorial of n1 using the fact_step constructor of the inductive relation. Finally, the factorial_equiv lemma is used to show that the new expression is equal to the value of the recursive function for n.




Together, these two lemmas show that the inductive relation and recursive function definitions of factorial are equivalent.




> Modify the factorial_req_equiv lemma to prove: forall n m, factorial_rec n = m > factorial n m




Certainly! Here's the modified lemma:




java




Lemma factorial_rec_equiv : forall n m, factorial_rec n = m > factorial n m.


Proof.


intros n m H.


rewrite < H.


clear H.


induction n.


 apply fact_base.


 simpl.


apply (fact_step n (factorial_rec n)).


apply IHn.


reflexivity.


Qed.




In this version of the lemma, we've added an extra argument m to the statement, which is the value that we want to prove factorial n equals to. We've also modified the proof to use the given m rather than the factorial of n as the target value.




We start by assuming factorial_rec n = m and then use the rewrite tactic to replace m with factorial_rec n in the goal. Next, we perform induction on n and apply the base case fact_base when n = 0. For the inductive step, we use the fact_step constructor of the inductive relation, where we apply the induction hypothesis IHn with the argument factorial_rec n. We also use the reflexivity tactic to show that the two occurrences of factorial_rec n are equal.




Overall, this lemma shows that if factorial_rec n = m for some n and m, then factorial n m. 