# IsarMathLib

## Proofs by humans, for humans, formally verified by Isabelle/ZF proof assistant

theory Fol1 imports ZF.Trancl
begin

Isabelle/ZF builds on the first order logic. Almost everything one would like to have in this area is covered in the standard Isabelle libraries. The material in this theory provides some lemmas that are missing or allow for a more readable proof style.

### Notions and lemmas in FOL

This section contains mostly shortcuts and workarounds that allow to use more readable coding style.

The next lemma serves as a workaround to problems with applying the definition of transitivity (of a relation) in our coding style (any attempt to do something like using trans_def puts Isabelle in an infinite loop).

lemma Fol1_L2:

assumes A1: $$\forall x y z.\ \langle x, y\rangle \in r \wedge \langle y, z\rangle \in r \longrightarrow \langle x, z\rangle \in r$$

shows $$\text{trans}(r)$$proof
from A1 have $$\forall x y z.\ \langle x, y\rangle \in r \longrightarrow \langle y, z\rangle \in r \longrightarrow \langle x, z\rangle \in r$$ using imp_conj
then show $$thesis$$ unfolding trans_def
qed

Another workaround for the problem of Isabelle simplifier looping when the transitivity definition is used.

lemma Fol1_L3:

assumes A1: $$\text{trans}(r)$$ and A2: $$\langle a,b\rangle \in r \wedge \langle b,c\rangle \in r$$

shows $$\langle a,c\rangle \in r$$proof
from A1 have $$\forall x y z.\ \langle x, y\rangle \in r \longrightarrow \langle y, z\rangle \in r \longrightarrow \langle x, z\rangle \in r$$ unfolding trans_def
with A2 show $$thesis$$ using imp_conj
qed

There is a problem with application of the definition of asymetry for relations. The next lemma is a workaround.

lemma Fol1_L4:

assumes A1: $$\text{antisym}(r)$$ and A2: $$\langle a,b\rangle \in r$$, $$\langle b,a\rangle \in r$$

shows $$a=b$$proof
from A1 have $$\forall x y.\ \langle x,y\rangle \in r \longrightarrow \langle y,x\rangle \in r \longrightarrow x=y$$ unfolding antisym_def
with A2 show $$a=b$$ using imp_conj
qed

The definition below implements a common idiom that states that (perhaps under some assumptions) exactly one of given three statements is true.

definition

$$\text{Exactly_1_of_3_holds} (p,q,r) \equiv$$ $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$

The next lemma allows to prove statements of the form $$\text{Exactly_1_of_3_holds} (p,q,r)$$.

lemma Fol1_L5:

assumes $$p\vee q\vee r$$ and $$p \longrightarrow \neg q \wedge \neg r$$ and $$q \longrightarrow \neg p \wedge \neg r$$ and $$r \longrightarrow \neg p \wedge \neg q$$

shows $$\text{Exactly_1_of_3_holds} (p,q,r)$$proof
from assms have $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$
then show $$\text{Exactly_1_of_3_holds} (p,q,r)$$ unfolding Exactly_1_of_3_holds_def
qed

If exactly one of $$p,q,r$$ holds and $$p$$ is not true, then $$q$$ or $$r$$.

lemma Fol1_L6:

assumes A1: $$\neg p$$ and A2: $$\text{Exactly_1_of_3_holds} (p,q,r)$$

shows $$q\vee r$$proof
from A2 have $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$ unfolding Exactly_1_of_3_holds_def
hence $$p \vee q \vee r$$
with A1 show $$q \vee r$$
qed

If exactly one of $$p,q,r$$ holds and $$q$$ is true, then $$r$$ can not be true.

lemma Fol1_L7:

assumes A1: $$q$$ and A2: $$\text{Exactly_1_of_3_holds} (p,q,r)$$

shows $$\neg r$$proof
from A2 have $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$ unfolding Exactly_1_of_3_holds_def
with A1 show $$\neg r$$
qed

The next lemma demonstrates an elegant form of the $$\text{Exactly_1_of_3_holds} (p,q,r)$$ predicate.

lemma Fol1_L8:

shows $$\text{Exactly_1_of_3_holds} (p,q,r) \longleftrightarrow (p\longleftrightarrow q\longleftrightarrow r) \wedge \neg (p\wedge q\wedge r)$$proof
assume $$\text{Exactly_1_of_3_holds} (p,q,r)$$
then have $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$ unfolding Exactly_1_of_3_holds_def
thus $$(p\longleftrightarrow q\longleftrightarrow r) \wedge \neg (p\wedge q\wedge r)$$
next
assume $$(p\longleftrightarrow q\longleftrightarrow r) \wedge \neg (p\wedge q\wedge r)$$
hence $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$
then show $$\text{Exactly_1_of_3_holds} (p,q,r)$$ unfolding Exactly_1_of_3_holds_def
qed

A property of the Exactly_1_of_3_holds predicate.

lemma Fol1_L8A:

assumes A1: $$\text{Exactly_1_of_3_holds} (p,q,r)$$

shows $$p \longleftrightarrow \neg (q \vee r)$$proof
from A1 have $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$ unfolding Exactly_1_of_3_holds_def
then show $$p \longleftrightarrow \neg (q \vee r)$$
qed

Exclusive or definition. There is one also defined in the standard Isabelle, denoted xor, but it relates to boolean values, which are sets. Here we define a logical functor.

definition

$$p \text{ Xor } q \equiv (p\vee q) \wedge \neg (p \wedge q)$$

The "exclusive or" is the same as negation of equivalence.

lemma Fol1_L9:

shows $$p \text{ Xor } q \longleftrightarrow \neg (p\longleftrightarrow q)$$ using Xor_def

Equivalence relations are symmetric.

lemma equiv_is_sym:

assumes A1: $$\text{equiv}(X,r)$$ and A2: $$\langle x,y\rangle \in r$$

shows $$\langle y,x\rangle \in r$$proof
from A1 have $$sym(r)$$ using equiv_def
then have $$\forall x y.\ \langle x,y\rangle \in r \longrightarrow \langle y,x\rangle \in r$$ unfolding sym_def
with A2 show $$\langle y,x\rangle \in r$$
qed
end
Definition of Exactly_1_of_3_holds: $$\text{Exactly_1_of_3_holds} (p,q,r) \equiv$$ $$(p\vee q\vee r) \wedge (p \longrightarrow \neg q \wedge \neg r) \wedge (q \longrightarrow \neg p \wedge \neg r) \wedge (r \longrightarrow \neg p \wedge \neg q)$$
Definition of Xor: $$p \text{ Xor } q \equiv (p\vee q) \wedge \neg (p \wedge q)$$