src/HOL/Isar_Examples/First_Order_Logic.thy
changeset 61935 6512e84cc9f5
parent 61758 df6258b7e53f
child 63585 f4a308fdf664
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/src/HOL/Isar_Examples/First_Order_Logic.thy	Sat Dec 26 19:27:46 2015 +0100
@@ -0,0 +1,160 @@
+(*  Title:      HOL/Isar_Examples/First_Order_Logic.thy
+    Author:     Makarius
+*)
+
+section \<open>A simple formulation of First-Order Logic\<close>
+
+text \<open>
+  The subsequent theory development illustrates single-sorted intuitionistic
+  first-order logic with equality, formulated within the Pure framework.
+\<close>
+
+theory First_Order_Logic
+imports Pure
+begin
+
+subsection \<open>Abstract syntax\<close>
+
+typedecl i
+typedecl o
+
+judgment Trueprop :: "o \<Rightarrow> prop"  ("_" 5)
+
+
+subsection \<open>Propositional logic\<close>
+
+axiomatization false :: o  ("\<bottom>")
+  where falseE [elim]: "\<bottom> \<Longrightarrow> A"
+
+
+axiomatization imp :: "o \<Rightarrow> o \<Rightarrow> o"  (infixr "\<longrightarrow>" 25)
+  where impI [intro]: "(A \<Longrightarrow> B) \<Longrightarrow> A \<longrightarrow> B"
+    and mp [dest]: "A \<longrightarrow> B \<Longrightarrow> A \<Longrightarrow> B"
+
+
+axiomatization conj :: "o \<Rightarrow> o \<Rightarrow> o"  (infixr "\<and>" 35)
+  where conjI [intro]: "A \<Longrightarrow> B \<Longrightarrow> A \<and> B"
+    and conjD1: "A \<and> B \<Longrightarrow> A"
+    and conjD2: "A \<and> B \<Longrightarrow> B"
+
+theorem conjE [elim]:
+  assumes "A \<and> B"
+  obtains A and B
+proof
+  from \<open>A \<and> B\<close> show A
+    by (rule conjD1)
+  from \<open>A \<and> B\<close> show B
+    by (rule conjD2)
+qed
+
+
+axiomatization disj :: "o \<Rightarrow> o \<Rightarrow> o"  (infixr "\<or>" 30)
+  where disjE [elim]: "A \<or> B \<Longrightarrow> (A \<Longrightarrow> C) \<Longrightarrow> (B \<Longrightarrow> C) \<Longrightarrow> C"
+    and disjI1 [intro]: "A \<Longrightarrow> A \<or> B"
+    and disjI2 [intro]: "B \<Longrightarrow> A \<or> B"
+
+
+definition true :: o  ("\<top>")
+  where "\<top> \<equiv> \<bottom> \<longrightarrow> \<bottom>"
+
+theorem trueI [intro]: \<top>
+  unfolding true_def ..
+
+
+definition not :: "o \<Rightarrow> o"  ("\<not> _" [40] 40)
+  where "\<not> A \<equiv> A \<longrightarrow> \<bottom>"
+
+theorem notI [intro]: "(A \<Longrightarrow> \<bottom>) \<Longrightarrow> \<not> A"
+  unfolding not_def ..
+
+theorem notE [elim]: "\<not> A \<Longrightarrow> A \<Longrightarrow> B"
+  unfolding not_def
+proof -
+  assume "A \<longrightarrow> \<bottom>" and A
+  then have \<bottom> ..
+  then show B ..
+qed
+
+
+definition iff :: "o \<Rightarrow> o \<Rightarrow> o"  (infixr "\<longleftrightarrow>" 25)
+  where "A \<longleftrightarrow> B \<equiv> (A \<longrightarrow> B) \<and> (B \<longrightarrow> A)"
+
+theorem iffI [intro]:
+  assumes "A \<Longrightarrow> B"
+    and "B \<Longrightarrow> A"
+  shows "A \<longleftrightarrow> B"
+  unfolding iff_def
+proof
+  from \<open>A \<Longrightarrow> B\<close> show "A \<longrightarrow> B" ..
+  from \<open>B \<Longrightarrow> A\<close> show "B \<longrightarrow> A" ..
+qed
+
+theorem iff1 [elim]:
+  assumes "A \<longleftrightarrow> B" and A
+  shows B
+proof -
+  from \<open>A \<longleftrightarrow> B\<close> have "(A \<longrightarrow> B) \<and> (B \<longrightarrow> A)"
+    unfolding iff_def .
+  then have "A \<longrightarrow> B" ..
+  from this and \<open>A\<close> show B ..
+qed
+
+theorem iff2 [elim]:
+  assumes "A \<longleftrightarrow> B" and B
+  shows A
+proof -
+  from \<open>A \<longleftrightarrow> B\<close> have "(A \<longrightarrow> B) \<and> (B \<longrightarrow> A)"
+    unfolding iff_def .
+  then have "B \<longrightarrow> A" ..
+  from this and \<open>B\<close> show A ..
+qed
+
+
+subsection \<open>Equality\<close>
+
+axiomatization equal :: "i \<Rightarrow> i \<Rightarrow> o"  (infixl "=" 50)
+  where refl [intro]: "x = x"
+    and subst: "x = y \<Longrightarrow> P x \<Longrightarrow> P y"
+
+theorem trans [trans]: "x = y \<Longrightarrow> y = z \<Longrightarrow> x = z"
+  by (rule subst)
+
+theorem sym [sym]: "x = y \<Longrightarrow> y = x"
+proof -
+  assume "x = y"
+  from this and refl show "y = x"
+    by (rule subst)
+qed
+
+
+subsection \<open>Quantifiers\<close>
+
+axiomatization All :: "(i \<Rightarrow> o) \<Rightarrow> o"  (binder "\<forall>" 10)
+  where allI [intro]: "(\<And>x. P x) \<Longrightarrow> \<forall>x. P x"
+    and allD [dest]: "\<forall>x. P x \<Longrightarrow> P a"
+
+axiomatization Ex :: "(i \<Rightarrow> o) \<Rightarrow> o"  (binder "\<exists>" 10)
+  where exI [intro]: "P a \<Longrightarrow> \<exists>x. P x"
+    and exE [elim]: "\<exists>x. P x \<Longrightarrow> (\<And>x. P x \<Longrightarrow> C) \<Longrightarrow> C"
+
+
+lemma "(\<exists>x. P (f x)) \<longrightarrow> (\<exists>y. P y)"
+proof
+  assume "\<exists>x. P (f x)"
+  then obtain x where "P (f x)" ..
+  then show "\<exists>y. P y" ..
+qed
+
+lemma "(\<exists>x. \<forall>y. R x y) \<longrightarrow> (\<forall>y. \<exists>x. R x y)"
+proof
+  assume "\<exists>x. \<forall>y. R x y"
+  then obtain x where "\<forall>y. R x y" ..
+  show "\<forall>y. \<exists>x. R x y"
+  proof
+    fix y
+    from \<open>\<forall>y. R x y\<close> have "R x y" ..
+    then show "\<exists>x. R x y" ..
+  qed
+qed
+
+end