Currently I have to learn Coq and don't know how to deal with an or :
As an example, as simple as it is, I don't see how to prove:
Theorem T0: x \/ ~x.
I would really appreciate it, if someone could help me.
For reference I use this cheat sheet.
Also an example of a proof I have in mind: Here for double negation:
Require Import Classical_Prop.
Parameters x: Prop.
Theorem T7: (~~x) -> x.
intro H.
apply NNPP.
exact H.
Qed.
You cannot prove it in "vanilla" Coq, because it is based on intuitionistic logic:
From a proof-theoretic perspective, intuitionistic logic is a restriction of classical logic in which the law of excluded middle and double negation elimination are not valid logical rules.
There are several ways you can deal with a situation like this.
Introduce the law of excluded middle as an axiom:
Axiom excluded_middle : forall P:Prop, P \/ ~ P.
There is no more need to prove anything after this point.
Introduce some axiom equivalent to the law of excluded middle and prove their equivalence. Here is just a few examples.
Double negation elimination is one such axiom:
Axiom dne : forall P:Prop, ~~P -> P.
Peirce's law is another example:
Axiom peirce : forall P Q:Prop, ((P -> Q) -> P) -> P.
Or use one of the De Morgan's laws:
Axiom de_morgan_and : forall P Q:Prop, ~(~P /\ ~Q) -> P \/ Q.
As others informed you, your tautology is not a tautology unless you assume classical logic. But since you're doing tautologies on decidable truth values, you could use bool instead of Prop. Then your tautology holds:
Require Import Bool.
Lemma how_about_bool: forall (p : bool), Is_true (p || negb p).
Proof.
now intros [|].
Qed.
