Theory Fol1

theory Fol1
imports Trancl
(* 
    This file is a part of IsarMathLib - 
    a library of formalized mathematics for Isabelle/Isar.

    Copyright (C) 2005 - 2008  Slawomir Kolodynski

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*)

header{*\isaheader{Fol1.thy}*}

theory Fol1 imports Trancl

begin

text{*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. *}


section{*Notions and lemmas in FOL*}

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

text{*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 @{text "using trans_def"} results up Isabelle in an 
  infinite loop). *}

lemma Fol1_L2: assumes 
  A1: "∀ x y z. ⟨x, y⟩ ∈ r ∧ ⟨y, z⟩ ∈ r --> ⟨x, z⟩ ∈ r"
  shows "trans(r)"
proof -
  from A1 have
    "∀ x y z. ⟨x, y⟩ ∈ r --> ⟨y, z⟩ ∈ r --> ⟨x, z⟩ ∈ r"
    using imp_conj by blast;
  then show ?thesis unfolding trans_def by blast;
qed;

text{*Another workaround for the problem of Isabelle simplifier looping when 
  the transitivity definition is used. *}

lemma Fol1_L3: assumes A1: "trans(r)" and A2: "⟨ a,b⟩ ∈ r  ∧ ⟨ b,c⟩ ∈ r"
  shows "⟨ a,c⟩ ∈ r"
proof -
  from A1 have  "∀x y z. ⟨x, y⟩ ∈ r --> ⟨y, z⟩ ∈ r --> ⟨x, z⟩ ∈ r"
   unfolding trans_def by blast
  with A2 show ?thesis using imp_conj by fast;
qed
  
text{*There is a problem with application of the definition of asymetry for
  relations. The next lemma is a workaround.*}

lemma Fol1_L4: 
  assumes A1: "antisym(r)" and A2: "⟨ a,b⟩ ∈ r"   "⟨ b,a⟩ ∈ r"  
  shows "a=b"
proof -
  from A1 have "∀ x y. ⟨ x,y⟩ ∈ r --> ⟨ y,x⟩ ∈ r --> x=y"
    unfolding antisym_def by blast
  with A2 show "a=b" using imp_conj by fast;
qed;

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

definition
  "Exactly_1_of_3_holds(p,q,r) ≡ 
  (p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)";

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

lemma Fol1_L5:
  assumes "p∨q∨r"
  and "p --> ¬q ∧ ¬r"
  and "q --> ¬p ∧ ¬r"
  and "r --> ¬p ∧ ¬q"
  shows "Exactly_1_of_3_holds(p,q,r)"
proof -;
  from assms have
    "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    by blast;
  then show "Exactly_1_of_3_holds (p,q,r)"
    unfolding Exactly_1_of_3_holds_def by fast;
qed;

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

lemma Fol1_L6: 
  assumes A1: "¬p" and A2: "Exactly_1_of_3_holds(p,q,r)" 
  shows "q∨r"
proof -
  from A2 have  
    "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    unfolding Exactly_1_of_3_holds_def by fast
  hence "p ∨ q ∨ r" by blast;
  with A1 show "q ∨ r" by simp;
qed;

text{*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: "Exactly_1_of_3_holds(p,q,r)"
  shows "¬r"
proof -
   from A2 have  
    "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    unfolding Exactly_1_of_3_holds_def by fast
  with A1 show "¬r" by blast;
qed;

text{* The next lemma demonstrates an elegant form of the 
  @{text "Exactly_1_of_3_holds(p,q,r)"} predicate. More on that
  at www.solcon.nl/mklooster/calc/calc-tri.html . *}

lemma Fol1_L8: 
  shows "Exactly_1_of_3_holds(p,q,r) <-> (p<->q<->r) ∧ ¬(p∧q∧r)"
proof
  assume "Exactly_1_of_3_holds(p,q,r)"
  then have 
    "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    unfolding Exactly_1_of_3_holds_def by fast
  thus "(p<->q<->r) ∧ ¬(p∧q∧r)" by blast;
next assume "(p<->q<->r) ∧ ¬(p∧q∧r)" 
  hence
    "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    by auto;
  then show "Exactly_1_of_3_holds(p,q,r)"
    unfolding Exactly_1_of_3_holds_def by fast
qed;

text{*A property of the @{text "Exactly_1_of_3_holds"} predicate.*}

lemma Fol1_L8A: assumes A1: "Exactly_1_of_3_holds(p,q,r)"
  shows "p <-> ¬(q ∨ r)"
proof -
  from A1 have "(p∨q∨r) ∧ (p --> ¬q ∧ ¬r) ∧ (q --> ¬p ∧ ¬r) ∧ (r --> ¬p ∧ ¬q)"
    unfolding Exactly_1_of_3_holds_def by fast
  then show "p <-> ¬(q ∨ r)" by blast;
qed;

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

definition
  Xor (infixl "Xor" 66) where
  "p Xor q ≡ (p∨q) ∧ ¬(p ∧ q)"

text{*The "exclusive or" is the same as negation of equivalence.*}

lemma Fol1_L9: shows "p Xor q <-> ¬(p<->q)"
  using Xor_def by auto;

text{*Equivalence relations are symmetric.*}

lemma equiv_is_sym: assumes A1: "equiv(X,r)" and A2: "⟨x,y⟩ ∈ r"
  shows  "⟨y,x⟩ ∈ r"
proof -
  from A1 have "sym(r)" using equiv_def by simp;
  then have "∀x y. ⟨x,y⟩ ∈ r --> ⟨y,x⟩ ∈ r"
    unfolding sym_def by fast
  with A2 show "⟨y,x⟩ ∈ r" by blast;
qed;

(* In Isabelle/ZF conjunction associates to the right!.
lemma test: assumes A1: "P" "Q∧R"
  shows "P∧Q∧R"
proof - 
  from A1 show "P∧Q∧R" by (rule one_more_conj);
qed;

*)
end