## Turing Machine Arrow Definition - turing-machines

Does anyone know the exact definition of '=>', '<=' and '<=>' in the context of Turing Machines? Googling failed to provide me with the answer!
To put it into context, here's a theorem / proof.
_
Theorem A language L is decidable <=> both L and L' are Turing-recognisable.
Proof: => is obvious. For <=, we have TM’s M1 and M2 that recognise L, L' respectively. Use them to build a TM M that runs M1 and M2 in parallel until one of them accepts (which must happen). If M1 accepts M accepts too; if M2 accepts, M rejects.

Theorem A language L is decidable if and only if both L and L' are Turing-recognisable.
Proof: L is decidable implies L and L' are Turing recognisable is obvious. For both L and L' are Turing-recognisable implies L is decidable, we have TM’s M1 and M2 that recognise L, L' respectively. Use them to build a TM M that runs M1 and M2 in parallel until one of them accepts (which must happen). If M1 accepts M accepts too; if M2 accepts, M rejects.
<==> means if and only if. A => B means A implies B. In if and only if proofs we prove double implication. In other words, in order to prove that A if and only if B then we prove that a implies B and B implies A.

## Related

### Transfering proof from Z to N in Coq

Is there a way in Coq in to prove a statement for integer numbers and the transfer statement in a semi-automatic way to naturals? For a concret example, take the following lemma: Lemma cancellation: forall a b c: nat, a > 0 -> a * b = a * c -> b = c. The statement is actually true in Z. In this case, it is easier to prove, because one can use subtraction to get a * (b - c) = 0 and then simplify a. But subtraction for naturals is capped, hence this would not work. Assume I can prove this for integers. Is there some tactic that one could use to derive the statement for naturals as well?

One solution is a tactic called zify that automatically turns a goal manipulating naturals into a goal manipulating integers (e.g., by inserting appropriate calls to Z.of_nat). This tactic is internally called by lia, but does not seem to be well documented. It is at least mentioned here. In your case, this would give the following. Require Import ZArith. (* The lemma stated in Z. *) Lemma cancellation: (forall a b c, a > 0 -> a * b = a * c -> b = c)%Z. Proof. (* your favorite proof of this result *) Admitted. (* The lemma stated in nat. *) Lemma cancellation_nat: forall a b c: nat, a > 0 -> a * b = a * c -> b = c. Proof. intros. zify. eauto using cancellation. Qed.

### Proof irrelevance in COQ

I am thinking about proof irrelevance in COQ. One provable statement says: If equality of a type is decidable then there can be only one proof for the equality statement, namely reflexivity. I wonder if its possible to construct types with more than one equality proof in COQ. Therefore I ask if the following construct is consistent? (*it is known that f=g is undecidable in COQ *) Definition f(n:nat) := n. Definition g(n:nat) := n+0. Axiom p1: f=g. Axiom p2: f=g. Axiom nonirrelevance:p1<>p2. What me puzzles here is the fact that by introducing p1 I made the equality f=g decidable and therefore it should only have one proof! What is my error in reasoning here? Is that all a pure COQ behaviour or is it similar in HOTT?

I think you're confusing several things. The provable statement you are speaking of can be found in https://coq.inria.fr/library/Coq.Logic.Eqdep_dec.html and is Theorem eq_proofs_unicity A (eq_dec : forall x y : A, x = y \/ x <> y) (x : A) : forall (y:A) (p1 p2:x = y), p1 = p2. Now what is quite interesting is the type of eq_dec. First of all, it doesn't even really ask for equality to be decidable, it just asks for it to be true or false which is way less stronger than {x = y} + {x <> y} Then notice that it doesn't ask this just for the x and y to prove the equality irrevelance of, it ask this property for all functions. So you would need to prove your contradiction that forall (f g : nat -> nat), f = g \/ f <> g which you cannot. p1 is just a proof that f = g \/ f <> g for your specific f and g. Notice though that if you could, it would just mean that there is no way to build a system in which you can compare functions and yet have multiple ways which are provably different to check them. Finally, for P to be undecidable only means that there is no constructible functions over {P} + {~P} yet, it doesn't mean that adding one as an axiom leads to a contradiction. Just adding that in case it wasn't clear.

### Pigeonhole proof without decidable equality or excluded middle

In Software Foundations IndProp.v one is asked to prove the pigeonhole principle, and one may use excluded middle, but it is mentioned that it is not strictly necessary. I've been trying to prove it without EM, but my brain seems to be wired classically. Q: How would one prove the theorem without using excluded middle? How should one generally approach proofs for types without decidable equality, where one can't easily reason by cases? I'd be very happy for a complete proof to look at, but please avoid posting it "in the clear", so as to not spoil the exercise in the Software Foundations course. The definition uses two inductive predicates, In and repeats. Require Import Lists.List. Import ListNotations. Section Pigeon. Variable (X:Type). Implicit Type (x:X). Fixpoint In x l : Prop := (*** In ***) match l with | nil => False | (x'::l') => x' = x \/ In x l' end. Hypothesis in_split : forall x l, In x l -> exists l1 l2, l = l1 ++ x :: l2. Hypothesis in_app: forall x l1 l2, In x (l1++l2) <-> In x l1 \/ In x l2. Inductive repeats : list X -> Prop := (*** repeats ***) repeats_hd l x : In x l -> repeats (x::l) | repeats_tl l x : repeats l -> repeats (x::l). Theorem pigeonhole_principle_NO_EM: (*** pigeonhole ***) forall l1 l2, length l2 < length l1 -> (* There are more pigeons than nests *) (forall x, In x l1 -> In x l2) -> (* All pigeons are in some nest *) repeats l1. (* Thus, some pigeons share nest *) Proof. (* ??? *)

I'll describe the thought process that led me to a solution, in case it helps. We may apply induction and it is straightforward to reduce to the case l1 = a::l1', l2 = a::l2'. Now l1' is a subset of a::l2'. My EM-trained intuition is that one of the following holds: a is in l1'. a is not in l1'. In the latter case, each element of l1' is in a::l2' but differs from a, and therefore must be in l2'. Thus l1' is a subset of l2', and we can apply the inductive hypothesis. Unfortunately if In is not decidable, the above can't be directly formalized. In particular if equality is not decidable for the given type, it's difficult to prove elements are unequal, and therefore difficult to prove a negative statement like ~(In a l1'). However, we wanted to use that negative statement to prove a positive one, namely forall x, In x l1' -> In x l2' By analogy, suppose we wanted to prove P \/ Q Q -> R ------ P \/ R The above intuitive argument is like starting from P \/ ~P, and using ~P -> Q -> R in the second case. We can use a direct proof to avoid EM. Quantifying over the list l1' makes this a bit more complicated, but still we can construct a direct proof using the following lemma, which can be proven by induction. Lemma split_or {X} (l : list X) (P Q : X -> Prop) : (forall x, In x l -> (P x \/ Q x)) -> (exists x, In x l /\ P x) \/ (forall x, In x l -> Q x). Finally note that the intuitive pigeonhole principle could also be formalized as the following way, which cannot be proven when the type has undecidable equality (note that it has a negative statement in the conclusion): Definition pigeon2 {X} : Prop := forall (l1 l2 : list X), length l2 < length l1 -> (exists x, In x l1 /\ ~(In x l2)) \/ repeats l1.

A possible constructive proof goes like this: We prove pigeonhole_principle_NO_EM by induction on l1. Only the non-empty case is possible because of the length constraint. So, assume l1 = x :: l1'. Now, check whether there is some element of l1' which is mapped by f : (forall x, In x l1 -> In x l2) to the same membership proof as x. If there is such an x' element, then it follows that x = x', therefore l1 repeats. If there is no such element, then we can get l2' by removing the element that x is mapped to from l2, and apply the induction hypothesis to l2' and the appropriate f' : forall x, In x l1' -> In x l2' function. That's it, but I note that actually formalizing this proof is not easy with the definitions given, because we need to prove heterogeneous or dependent equalities, since we have to compare membership proofs for possibly different elements. As to the question of getting the hang of constructive proofs in general, an important skill or habit is always examining what kind of data we have, not just what kind of logical facts we know. In this case, membership proofs are actually indices pointing into lists, bundled together with proofs that the pointed-to elements equal certain values. If membership proofs didn't tell where exactly elements are located then this proof would not be possible constructively.

### How to introduce a new variable in Coq?

I was wondering if there is a way to introduce an entirely new variable during the proof of a theorem in Coq? For a complete example, consider the following property from here about the evenness of the length of a list. Inductive ev_list {X:Type}: list X -> Prop := | el_nil : ev_list [] | el_cc : forall x y l, ev_list l -> ev_list (x :: y :: l). Now I want to prove that for any list l if its length is even, then ev_list l holds: Lemma ev_length__ev_list': forall X (l : list X), ev (length l) -> ev_list l. Proof. intros X l H. which gives: 1 subgoals X : Type l : list X H : ev (length l) ______________________________________(1/1) ev_list l Now, I'd like to "define" a new free variable n and a hypothesis n = length l. In hand-written math, I think we can do this, and then do induction about n. But is there a way to do the same in Coq? Note. the reasons I ask are that: I don't want to introduce this n artificially into the statement of the theorem itself, as is done in the page linked earlier, which IMHO is unnatural. I tried to induction H., but it seems not working. Coq wasn't able to do case analysis on length l's ev-ness, and no induction hypothesis (IH) was generated. Thanks.

This is a common issue in Coq proofs. You can use the remember tactic: remember (length l) as n. If you're doing induction on H as well, you might also have to generalize over l beforehand, by doing generalize dependent l. induction H.

If you want to add a new variable only for your induction, you can use directly induction (length l) eqn:H0

According to the Curry-Howard Isomorphism, hypothesis in your context are just variables. You can define new variables with a function. The following refine tactic extends the goal with a fresh variable n (that is set to length l) and a proof e that n = length l (that is set to eq_refl). Lemma ev_length__ev_list': forall X (l : list X), ev (length l) -> ev_list l. Proof. intros X l H. refine ((fun n (e:n = length l) => _) (length l) eq_refl). (* proof *) Admitted.

### How to create a new hypothesis from apply?

When I run the Coq script below (a simplification of the original one): Inductive w (g: nat): nat -> Prop:= | z: w g 0. Lemma x: forall (i j: nat), w i j -> (forall k: nat, k <= k). Proof. Admitted. Lemma y: forall (m n: nat), w m n -> w m n. Proof. intros m n H. apply x in H. I get the following error message on the last line: Error: Unable to find an instance for the variable k. Can anybody explain to me why this happens and what I have to do in order to have forall k: nat, k <= k as a new hypothesis in the context? Thanks in advance, Marcus.

Rather than apply x in H, you can use pose proof (x _ _ H). This will give you the hypothesis you're looking for. From the Coq tactic manual, apply term in ident tries to match the conclusion of the type of ident against a non-dependent premise of the type of term, trying them from right to left. I think the key point to note here is that apply only works for non-dependent premises, while the premise you want it to match, i=j, is dependent. However, the particular error message returned by Coq is confusing.

Since your lemma x is contains an inner universal quantification (the forall k part at the end), Coq does not managed to guess which natural number you want to use. By applying xto H, you only provide i and j. You have two solutions: provide the relevant k by hand using the apply x with (k := foo) in H syntax ask Coq to introduce a "meta-variable" (think of it as a typed hole that you will fill later) using the eapply tactic. Hope it helps, V.