definition (*composition of relations and functions; NOT Suppes's relative product*)
comp :: "[i,i]→i" (infixr‹O›60) where "r O s ≡ {xz ∈ domain(s)*range(r) . ∃x y z. xz=⟨x,z⟩∧⟨x,y⟩:s ∧⟨y,z⟩:r}"
definition (*the identity function for A*)
id :: "i→i"where "id(A) ≡ (λx∈A. x)"
definition (*one-to-one functions from A to B*)
inj :: "[i,i]→i"where "inj(A,B) ≡ { f ∈ A->B. ∀w∈A. ∀x∈A. f`w=f`x ⟶ w=x}"
definition (*onto functions from A to B*)
surj :: "[i,i]→i"where "surj(A,B) ≡ { f ∈ A->B . ∀y∈B. ∃x∈A. f`x=y}"
definition (*one-to-one and onto functions*)
bij :: "[i,i]→i"where "bij(A,B) ≡ inj(A,B) ∩ surj(A,B)"
text‹A function with a right inverse is a surjection›
lemma f_imp_surjective: "[f ∈ A->B; ∧y. y ∈ B ==> d(y): A; ∧y. y ∈ B ==> f`d(y) = y] ==> f ∈ surj(A,B)" by (simp add: surj_def, blast)
lemma lam_surjective: "[∧x. x ∈ A ==> c(x): B; ∧y. y ∈ B ==> d(y): A; ∧y. y ∈ B ==> c(d(y)) = y \<rbrakk> ==> (λx∈A. c(x)) ∈ surj(A,B)" apply (rule_tac d = d in f_imp_surjective) apply (simp_all add: lam_type) done
text‹Good for dealing with sets of pairs, but a bit ugly in use [used in AC]› lemma inj_equality: "[⟨a,b⟩:f; ⟨c,b⟩:f; f ∈ inj(A,B)]==> a=c" unfolding inj_def apply (blast dest: Pair_mem_PiD) done
lemma inj_apply_equality: "[f ∈ inj(A,B); f`a=f`b; a ∈ A; b ∈ A]==> a=b" by (unfold inj_def, blast)
text‹A function with a left inverse is an injection›
lemma lam_injective: "[∧x. x ∈ A ==> c(x): B; ∧x. x ∈ A ==> d(c(x)) = x] ==> (λx∈A. c(x)) ∈ inj(A,B)" apply (rule_tac d = d in f_imp_injective) apply (simp_all add: lam_type) done
text‹The premises are equivalent to saying that f is injective...› lemma left_inverse_lemma: "[f ∈ A->B; converse(f): C->A; a ∈ A]==> converse(f)`(f`a) = a" by (blast intro: apply_Pair apply_equality converseI)
lemma left_inverse [simp]: "[f ∈ inj(A,B); a ∈ A]==> converse(f)`(f`a) = a" by (blast intro: left_inverse_lemma inj_converse_fun inj_is_fun)
lemma left_inverse_eq: "[f ∈ inj(A,B); f ` x = y; x ∈ A]==> converse(f) ` y = x" by auto
(*Should the premises be f \<in> surj(A,B), b \<in> B for symmetry with left_inverse?
No: they would not imply that converse(f) was a function! *) lemma right_inverse [simp]: "[f ∈ inj(A,B); b ∈ range(f)]==> f`(converse(f)`b) = b" by (blast intro: right_inverse_lemma inj_converse_fun inj_is_fun)
lemma right_inverse_bij: "[f ∈ bij(A,B); b ∈ B]==> f`(converse(f)`b) = b" by (force simp add: bij_def surj_range)
subsection‹Converses of Injections, Surjections, Bijections›
text‹Adding this as an intro! rule seems to cause looping› lemma bij_converse_bij [TC]: "f ∈ bij(A,B) ==> converse(f): bij(B,A)" unfolding bij_def apply (fast elim: surj_range [THEN subst] inj_converse_inj inj_converse_surj) done
subsection‹Composition of Two Relations›
text‹The inductive definition package could derive these theorems for term‹r O s››
lemma compI [intro]: "[⟨a,b⟩:s; ⟨b,c⟩:r]==>⟨a,c⟩∈ r O s" by (unfold comp_def, blast)
lemma compE [elim!]: "[xz ∈ r O s; ∧x y z. [xz=⟨x,z⟩; ⟨x,y⟩:s; ⟨y,z⟩:r]==> P] ==> P" by (unfold comp_def, blast)
lemma compEpair: "[⟨a,c⟩∈ r O s; ∧y. [⟨a,y⟩:s; ⟨y,c⟩:r]==> P] ==> P" by (erule compE, simp)
lemma converse_comp: "converse(R O S) = converse(S) O converse(R)" by blast
subsection‹Domain and Range -- see Suppes, Section 3.1›
text‹Boyer et al., Set Theory in First-Order Logic, JAR 2 (1986), 287-327› lemma range_comp: "range(r O s) ⊆ range(r)" by blast
lemma range_comp_eq: "domain(r) ⊆ range(s) ==> range(r O s) = range(r)" by (rule range_comp [THEN equalityI], blast)
lemma domain_comp: "domain(r O s) ⊆ domain(s)" by blast
lemma domain_comp_eq: "range(s) ⊆ domain(r) ==> domain(r O s) = domain(s)" by (rule domain_comp [THEN equalityI], blast)
lemma image_comp: "(r O s)``A = r``(s``A)" by blast
lemma inj_inj_range: "f ∈ inj(A,B) ==> f ∈ inj(A,range(f))" by (auto simp add: inj_def Pi_iff function_def)
lemma inj_bij_range: "f ∈ inj(A,B) ==> f ∈ bij(A,range(f))" by (auto simp add: bij_def intro: inj_inj_range inj_is_fun fun_is_surj)
subsection‹Other Results›
lemma comp_mono: "[r'<=r; s'<=s]==> (r' O s') ⊆ (r O s)" by blast
text‹composition preserves relations› lemma comp_rel: "[s<=A*B; r<=B*C]==> (r O s) ⊆ A*C" by blast
text‹associative law for composition› lemma comp_assoc: "(r O s) O t = r O (s O t)" by blast
(*left identity of composition; provable inclusions are id(A)Or\<subseteq>r
and \<lbrakk>r<=A*B; B<=C\<rbrakk> \<Longrightarrow> r \<subseteq> id(C) O r *) lemma left_comp_id: "r<=A*B ==> id(B) O r = r" by blast
(*right identity of composition; provable inclusions are rOid(A)\<subseteq>r
and \<lbrakk>r<=A*B; A<=C\<rbrakk> \<Longrightarrow> r \<subseteq> r O id(C) *) lemma right_comp_id: "r<=A*B ==> r O id(A) = r" by blast
subsection‹Composition Preserves Functions, Injections, and Surjections›
lemma comp_function: "[function(g); function(f)]==> function(f O g)" by (unfold function_def, blast)
text‹Don't think the premises can be weakened much› lemma comp_fun: "[g ∈ A->B; f ∈ B->C]==> (f O g) ∈ A->C" apply (auto simp add: Pi_def comp_function Pow_iff comp_rel) apply (subst range_rel_subset [THEN domain_comp_eq], auto) done
(*Thanks to the new definition of "apply", the premise f \<in> B->C is gone!*) lemma comp_fun_apply [simp]: "[g ∈ A->B; a ∈ A]==> (f O g)`a = f`(g`a)" apply (frule apply_Pair, assumption) apply (simp add: apply_def image_comp) apply (blast dest: apply_equality) done
text‹Simplifies compositions of lambda-abstractions› lemma comp_lam: "[∧x. x ∈ A ==> b(x): B] ==> (λy∈B. c(y)) O (λx∈A. b(x)) = (λx∈A. c(b(x)))" apply (subgoal_tac "(λx∈A. b(x)) ∈ A -> B") apply (rule fun_extension) apply (blast intro: comp_fun lam_funtype) apply (rule lam_funtype) apply simp apply (simp add: lam_type) done
subsection‹Dual Properties of term‹inj› and term‹surj››
text‹Useful for proofs from
D Pastre. Automatic theorem proving in set theory.
Artificial Intelligence, 10:1--27, 1978.›
lemma comp_mem_injD1: "[(f O g): inj(A,C); g ∈ A->B; f ∈ B->C]==> g ∈ inj(A,B)" by (unfold inj_def, force)
lemma comp_mem_injD2: "[(f O g): inj(A,C); g ∈ surj(A,B); f ∈ B->C]==> f ∈ inj(B,C)" apply (unfold inj_def surj_def, safe) apply (rule_tac x1 = x in bspec [THEN bexE]) apply (erule_tac [3] x1 = w in bspec [THEN bexE], assumption+, safe) apply (rule_tac t = "(`) (g) "in subst_context) apply (erule asm_rl bspec [THEN bspec, THEN mp])+ apply (simp (no_asm_simp)) done
lemma comp_mem_surjD1: "[(f O g): surj(A,C); g ∈ A->B; f ∈ B->C]==> f ∈ surj(B,C)" unfolding surj_def apply (blast intro!: comp_fun_apply [symmetric] apply_funtype) done
lemma comp_mem_surjD2: "[(f O g): surj(A,C); g ∈ A->B; f ∈ inj(B,C)]==> g ∈ surj(A,B)" apply (unfold inj_def surj_def, safe) apply (drule_tac x = "f`y"in bspec, auto) apply (blast intro: apply_funtype) done
subsubsection‹Inverses of Composition›
text‹left inverse of composition; one inclusion is term‹f ∈ A->B ==> id(A) ⊆ converse(f) O f›› lemma left_comp_inverse: "f ∈ inj(A,B) ==> converse(f) O f = id(A)" apply (unfold inj_def, clarify) apply (rule equalityI) apply (auto simp add: apply_iff, blast) done
text‹right inverse of composition; one inclusion is term‹f ∈ A->B ==> f O converse(f) ⊆ id(B)›› lemma right_comp_inverse: "f ∈ surj(A,B) ==> f O converse(f) = id(B)" apply (simp add: surj_def, clarify) apply (rule equalityI) apply (best elim: domain_type range_type dest: apply_equality2) apply (blast intro: apply_Pair) done
subsubsection‹Proving that a Function is a Bijection›
lemma comp_eq_id_iff: "[f ∈ A->B; g ∈ B->A]==> f O g = id(B) ⟷ (∀y∈B. f`(g`y)=y)" apply (unfold id_def, safe) apply (drule_tac t = "λh. h`y "in subst_context) apply simp apply (rule fun_extension) apply (blast intro: comp_fun lam_type) apply auto done
lemma fg_imp_bijective: "[f ∈ A->B; g ∈ B->A; f O g = id(B); g O f = id(A)]==> f ∈ bij(A,B)" unfolding bij_def apply (simp add: comp_eq_id_iff) apply (blast intro: f_imp_injective f_imp_surjective apply_funtype) done
lemma nilpotent_imp_bijective: "[f ∈ A->A; f O f = id(A)]==> f ∈ bij(A,A)" by (blast intro: fg_imp_bijective)
lemma invertible_imp_bijective: "[converse(f): B->A; f ∈ A->B]==> f ∈ bij(A,B)" by (simp add: fg_imp_bijective comp_eq_id_iff
left_inverse_lemma right_inverse_lemma)
subsubsection‹Unions of Functions›
text‹See similar theorems in func.thy›
text‹Theorem by KG, proof by LCP› lemma inj_disjoint_Un: "[f ∈ inj(A,B); g ∈ inj(C,D); B ∩ D = 0] ==> (λa∈A ∪ C. if a ∈ A then f`a else g`a) ∈ inj(A ∪ C, B ∪ D)" apply (rule_tac d = "λz. if z ∈ B then converse (f) `z else converse (g) `z" in lam_injective) apply (auto simp add: inj_is_fun [THEN apply_type]) done
lemma surj_disjoint_Un: "[f ∈ surj(A,B); g ∈ surj(C,D); A ∩ C = 0] ==> (f ∪ g) ∈ surj(A ∪ C, B ∪ D)" apply (simp add: surj_def fun_disjoint_Un) apply (blast dest!: domain_of_fun
intro!: fun_disjoint_apply1 fun_disjoint_apply2) done
text‹A simple, high-level proof; the version for injections follows from it,
using term‹f ∈ inj(A,B) ⟷ f ∈ bij(A,range(f))›› lemma bij_disjoint_Un: "[f ∈ bij(A,B); g ∈ bij(C,D); A ∩ C = 0; B ∩ D = 0] ==> (f ∪ g) ∈ bij(A ∪ C, B ∪ D)" apply (rule invertible_imp_bijective) apply (subst converse_Un) apply (auto intro: fun_disjoint_Un bij_is_fun bij_converse_bij) done
subsubsection‹Restrictions as Surjections and Bijections›
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