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 \ prop" (\_\ 5)
subsection \<open>Propositional logic\<close>
axiomatization false :: o (\<open>\<bottom>\<close>) where falseE [elim]: "\ \ A"
axiomatization imp :: "o \ o \ o" (infixr \\\ 25) where impI [intro]: "(A \ B) \ A \ B" and mp [dest]: "A \ B \ A \ B"
axiomatization conj :: "o \ o \ o" (infixr \\\ 35) where conjI [intro]: "A \ B \ A \ B" and conjD1: "A \ B \ A" and conjD2: "A \ B \ B"
theorem conjE [elim]: assumes"A \ 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 \ o \ o" (infixr \\\ 30) where disjE [elim]: "A \ B \ (A \ C) \ (B \ C) \ C" and disjI1 [intro]: "A \ A \ B" and disjI2 [intro]: "B \ A \ B"
definition true :: o (\<open>\<top>\<close>) where"\ \ \ \ \"
theoremnotE [elim]: "\ A \ A \ B" unfolding not_def proof - assume"A \ \" and A thenhave\<bottom> .. thenshow B .. qed
definition iff :: "o \ o \ o" (infixr \\\ 25) where"A \ B \ (A \ B) \ (B \ A)"
theorem iffI [intro]: assumes"A \ B" and"B \ A" shows"A \ 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 \ B" and A shows B proof - from\<open>A \<longleftrightarrow> B\<close> have "(A \<longrightarrow> B) \<and> (B \<longrightarrow> A)" unfolding iff_def . thenhave"A \ B" .. from this and\<open>A\<close> show B .. qed
theorem iff2 [elim]: assumes"A \ B" and B shows A proof - from\<open>A \<longleftrightarrow> B\<close> have "(A \<longrightarrow> B) \<and> (B \<longrightarrow> A)" unfolding iff_def . thenhave"B \ A" .. from this and\<open>B\<close> show A .. qed
subsection \<open>Equality\<close>
axiomatization equal :: "i \ i \ o" (infixl \=\ 50) where refl [intro]: "x = x" and subst: "x = y \ P x \ P y"
theorem trans [trans]: "x = y \ y = z \ x = z" by (rule subst)
theorem sym [sym]: "x = y \ y = x" proof - assume"x = y" from this and refl show"y = x" by (rule subst) qed
subsection \<open>Quantifiers\<close>
axiomatization All :: "(i \ o) \ o" (binder \\\ 10) where allI [intro]: "(\x. P x) \ \x. P x" and allD [dest]: "\x. P x \ P a"
axiomatization Ex :: "(i \ o) \ o" (binder \\\ 10) where exI [intro]: "P a \ \x. P x" and exE [elim]: "\x. P x \ (\x. P x \ C) \ C"
lemma"(\x. P (f x)) \ (\y. P y)" proof assume"\x. P (f x)" thenobtain x where"P (f x)" .. thenshow"\y. P y" .. qed
lemma"(\x. \y. R x y) \ (\y. \x. R x y)" proof assume"\x. \y. R x y" thenobtain x where"\y. R x y" .. show"\y. \x. R x y" proof fix y from\<open>\<forall>y. R x y\<close> have "R x y" .. thenshow"\x. R x y" .. qed qed
end
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