Theory Group_ZF_1

(* 
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section ‹Groups 1›

theory Group_ZF_1 imports Group_ZF

begin

text‹In this theory we consider right and left translations and odd 
  functions.›

subsection‹Translations›

text‹In this section we consider translations. Translations are maps 
  $T: G\rightarrow G$ of the form $T_g (a) = g\cdot a$ or 
  $T_g (a) = a\cdot g$. We also consider two-dimensional translations
  $T_g : G\times G \rightarrow G\times G$, where 
  $T_g(a,b) = (a\cdot g, b\cdot g)$ or $T_g(a,b) = (g\cdot a, g\cdot b)$. 
›

text‹For an element $a\in G$ the right translation is defined 
  a function (set of pairs) such that its value (the second element
  of a pair) is the value of the group operation on the first element
  of the pair and $g$. This looks a bit strange in the raw set notation, 
  when we write a function explicitely as a set of pairs and value of 
  the group operation on the pair $\langle a,b \rangle$ 
  as ‹P`⟨a,b⟩› instead of the usual infix $a\cdot b$
  or $a + b$.›

definition
  "RightTranslation(G,P,g) ≡ {⟨ a,b⟩ ∈ G×G. P`⟨a,g⟩ = b}"

text‹A similar definition of the left translation.›

definition
  "LeftTranslation(G,P,g) ≡ {⟨a,b⟩ ∈ G×G. P`⟨g,a⟩ = b}"

text‹Translations map $G$ into $G$. Two dimensional translations map
  $G\times G$ into itself.›

lemma (in group0) group0_5_L1: assumes A1: "g∈G"
  shows "RightTranslation(G,P,g) : G→G" and  "LeftTranslation(G,P,g) : G→G"
proof -
  from A1 have "∀a∈G. a⋅g ∈ G" and "∀a∈G. g⋅a ∈ G"
    using group_oper_assocA apply_funtype by auto
  then show 
    "RightTranslation(G,P,g) : G→G" 
    "LeftTranslation(G,P,g) : G→G"
    using RightTranslation_def LeftTranslation_def func1_1_L11A
    by auto
qed

text‹The values of the translations are what we expect.›

lemma (in group0) group0_5_L2: assumes "g∈G" "a∈G"
  shows 
  "RightTranslation(G,P,g)`(a) = a⋅g"
  "LeftTranslation(G,P,g)`(a) = g⋅a"
  using assms group0_5_L1 RightTranslation_def LeftTranslation_def
    func1_1_L11B by auto

text‹Composition of left translations is a left translation by the product.›

lemma (in group0) group0_5_L4: assumes A1: "g∈G" "h∈G" "a∈G" and 
  A2: "T⇩g = LeftTranslation(G,P,g)"  "T⇩h = LeftTranslation(G,P,h)"
  shows 
  "T⇩g`(T⇩h`(a)) = g⋅h⋅a"
  "T⇩g`(T⇩h`(a)) = LeftTranslation(G,P,g⋅h)`(a)"
proof -
  from A1 have I: "h⋅a∈G"  "g⋅h∈G"
    using group_oper_assocA apply_funtype by auto
  with A1 A2 show "T⇩g`(T⇩h`(a)) = g⋅h⋅a"
    using group0_5_L2 group_oper_assoc by simp
  with A1 A2 I show 
    "T⇩g`(T⇩h`(a)) = LeftTranslation(G,P,g⋅h)`(a)"
    using group0_5_L2 group_oper_assoc by simp
qed


text‹Composition of right translations is a right translation by 
  the product.›

lemma (in group0) group0_5_L5: assumes A1: "g∈G" "h∈G" "a∈G" and 
  A2: "T⇩g = RightTranslation(G,P,g)"  "T⇩h = RightTranslation(G,P,h)"
  shows 
 "T⇩g`(T⇩h`(a)) = a⋅h⋅g"
  "T⇩g`(T⇩h`(a)) = RightTranslation(G,P,h⋅g)`(a)"
proof -
  from A1 have I: "a⋅h∈G" "h⋅g ∈G"
    using group_oper_assocA apply_funtype by auto
  with A1 A2 show "T⇩g`(T⇩h`(a)) = a⋅h⋅g" 
    using group0_5_L2 group_oper_assoc by simp
  with A1 A2 I show 
    "T⇩g`(T⇩h`(a)) = RightTranslation(G,P,h⋅g)`(a)"
    using group0_5_L2 group_oper_assoc by simp
qed

text‹Point free version of ‹group0_5_L4› and ‹group0_5_L5›.›

lemma (in group0) trans_comp: assumes "g∈G" "h∈G" shows
  "RightTranslation(G,P,g) O RightTranslation(G,P,h) = RightTranslation(G,P,h⋅g)"
  "LeftTranslation(G,P,g) O LeftTranslation(G,P,h) = LeftTranslation(G,P,g⋅h)"
proof -
  let ?T⇩g = "RightTranslation(G,P,g)"
  let ?T⇩h = "RightTranslation(G,P,h)"
  from assms have "?T⇩g:G→G" and "?T⇩h:G→G"
    using group0_5_L1 by auto
  then have "?T⇩g O ?T⇩h:G→G" using comp_fun by simp
  moreover from assms have "RightTranslation(G,P,h⋅g):G→G"
    using group_op_closed group0_5_L1 by simp
  moreover from assms ‹?T⇩h:G→G› have 
    "∀a∈G. (?T⇩g O ?T⇩h)`(a) = RightTranslation(G,P,h⋅g)`(a)"
    using comp_fun_apply group0_5_L5 by simp
  ultimately show "?T⇩g O ?T⇩h =  RightTranslation(G,P,h⋅g)"
    by (rule func_eq)
next
  let ?T⇩g = "LeftTranslation(G,P,g)"
  let ?T⇩h = "LeftTranslation(G,P,h)"
  from assms have "?T⇩g:G→G" and "?T⇩h:G→G"
    using group0_5_L1 by auto
  then have "?T⇩g O ?T⇩h:G→G" using comp_fun by simp
  moreover from assms have "LeftTranslation(G,P,g⋅h):G→G"
    using group_op_closed group0_5_L1 by simp
  moreover from assms ‹?T⇩h:G→G› have 
    "∀a∈G. (?T⇩g O ?T⇩h)`(a) = LeftTranslation(G,P,g⋅h)`(a)"
    using comp_fun_apply group0_5_L4 by simp
  ultimately show "?T⇩g O ?T⇩h =  LeftTranslation(G,P,g⋅h)"
    by (rule func_eq)
qed

text‹The image of a set under a composition of translations is the same as
  the image under translation by a product.›

lemma (in group0) trans_comp_image: assumes A1: "g∈G" "h∈G" and
  A2: "T⇩g = LeftTranslation(G,P,g)"  "T⇩h = LeftTranslation(G,P,h)"
shows "T⇩g``(T⇩h``(A)) = LeftTranslation(G,P,g⋅h)``(A)"
proof -
  from A2 have "T⇩g``(T⇩h``(A)) = (T⇩g O T⇩h)``(A)"
    using image_comp by simp
  with assms show ?thesis using trans_comp by simp
qed

text‹Another form of the image of a set under a composition of translations›

lemma (in group0) group0_5_L6: 
  assumes A1: "g∈G" "h∈G" and A2: "A⊆G" and 
  A3: "T⇩g = RightTranslation(G,P,g)"  "T⇩h = RightTranslation(G,P,h)"
  shows "T⇩g``(T⇩h``(A)) = {a⋅h⋅g. a∈A}"
proof -
  from A2 have "∀a∈A. a∈G" by auto
  from A1 A3 have "T⇩g : G→G"  "T⇩h : G→G"
    using group0_5_L1 by auto
  with assms ‹∀a∈A. a∈G› show 
    "T⇩g``(T⇩h``(A)) = {a⋅h⋅g. a∈A}"
    using func1_1_L15C group0_5_L5 by auto
qed

text‹The translation by neutral element is the identity on group.›

lemma (in group0) trans_neutral: shows 
  "RightTranslation(G,P,𝟭) = id(G)" and "LeftTranslation(G,P,𝟭) = id(G)"
proof -
  have "RightTranslation(G,P,𝟭):G→G" and "∀a∈G. RightTranslation(G,P,𝟭)`(a) = a"
    using group0_2_L2 group0_5_L1 group0_5_L2  by auto
  then show "RightTranslation(G,P,𝟭) = id(G)" by (rule indentity_fun)
  have "LeftTranslation(G,P,𝟭):G→G" and "∀a∈G. LeftTranslation(G,P,𝟭)`(a) = a"
    using group0_2_L2 group0_5_L1 group0_5_L2  by auto
  then show "LeftTranslation(G,P,𝟭) = id(G)" by (rule indentity_fun)
qed

text‹Composition of translations by an element and its inverse is identity.›

lemma (in group0) trans_comp_id: assumes "g∈G" shows
  "RightTranslation(G,P,g) O RightTranslation(G,P,g¯) = id(G)" and
  "RightTranslation(G,P,g¯) O RightTranslation(G,P,g) = id(G)" and
  "LeftTranslation(G,P,g) O LeftTranslation(G,P,g¯) = id(G)" and
  "LeftTranslation(G,P,g¯) O LeftTranslation(G,P,g) = id(G)"
  using assms inverse_in_group trans_comp group0_2_L6 trans_neutral by auto

text‹Translations are bijective.›

lemma (in group0) trans_bij: assumes "g∈G" shows
  "RightTranslation(G,P,g) ∈ bij(G,G)" and "LeftTranslation(G,P,g) ∈ bij(G,G)"
proof-
  from assms have 
    "RightTranslation(G,P,g):G→G" and
    "RightTranslation(G,P,g¯):G→G" and
    "RightTranslation(G,P,g) O RightTranslation(G,P,g¯) = id(G)"
    "RightTranslation(G,P,g¯) O RightTranslation(G,P,g) = id(G)"
  using inverse_in_group group0_5_L1 trans_comp_id by auto
  then show "RightTranslation(G,P,g) ∈ bij(G,G)" using fg_imp_bijective by simp
  from assms have 
    "LeftTranslation(G,P,g):G→G" and
    "LeftTranslation(G,P,g¯):G→G" and
    "LeftTranslation(G,P,g) O LeftTranslation(G,P,g¯) = id(G)"
    "LeftTranslation(G,P,g¯) O LeftTranslation(G,P,g) = id(G)"
    using inverse_in_group group0_5_L1 trans_comp_id by auto
  then show "LeftTranslation(G,P,g) ∈ bij(G,G)" using fg_imp_bijective by simp
qed

text‹Converse of a translation is translation by the inverse.›

lemma (in group0) trans_conv_inv: assumes "g∈G" shows
  "converse(RightTranslation(G,P,g)) = RightTranslation(G,P,g¯)" and
  "converse(LeftTranslation(G,P,g)) = LeftTranslation(G,P,g¯)" and
  "LeftTranslation(G,P,g) = converse(LeftTranslation(G,P,g¯))" and
  "RightTranslation(G,P,g) = converse(RightTranslation(G,P,g¯))"
proof -
  from assms have
    "RightTranslation(G,P,g) ∈ bij(G,G)"  "RightTranslation(G,P,g¯) ∈ bij(G,G)" and
    "LeftTranslation(G,P,g) ∈ bij(G,G)"  "LeftTranslation(G,P,g¯) ∈ bij(G,G)"
    using trans_bij inverse_in_group by auto
  moreover from assms have
    "RightTranslation(G,P,g¯) O RightTranslation(G,P,g) = id(G)" and
    "LeftTranslation(G,P,g¯) O LeftTranslation(G,P,g) = id(G)" and
    "LeftTranslation(G,P,g) O LeftTranslation(G,P,g¯) = id(G)" and
    "LeftTranslation(G,P,g¯) O LeftTranslation(G,P,g) = id(G)"
    using trans_comp_id by auto
  ultimately show 
    "converse(RightTranslation(G,P,g)) = RightTranslation(G,P,g¯)" and
    "converse(LeftTranslation(G,P,g)) = LeftTranslation(G,P,g¯)" and  
    "LeftTranslation(G,P,g) = converse(LeftTranslation(G,P,g¯))" and
    "RightTranslation(G,P,g) = converse(RightTranslation(G,P,g¯))"
    using comp_id_conv by auto
qed
  
text‹The image of a set by translation is the same as the inverse image by
by the inverse element translation.›

lemma (in group0) trans_image_vimage: assumes "g∈G" shows
  "LeftTranslation(G,P,g)``(A) = LeftTranslation(G,P,g¯)-``(A)" and
  "RightTranslation(G,P,g)``(A) = RightTranslation(G,P,g¯)-``(A)"
  using assms trans_conv_inv vimage_converse by auto

text‹Another way of looking at translations is that they are sections
  of the group operation.›

lemma (in group0) trans_eq_section: assumes "g∈G" shows
  "RightTranslation(G,P,g) = Fix2ndVar(P,g)" and
  "LeftTranslation(G,P,g) =  Fix1stVar(P,g)"
proof -
  let ?T = "RightTranslation(G,P,g)"
  let ?F = "Fix2ndVar(P,g)"
  from assms have "?T: G→G" and "?F: G→G"
    using group0_5_L1 group_oper_assocA fix_2nd_var_fun by auto
  moreover from assms have "∀a∈G. ?T`(a) = ?F`(a)"
    using group0_5_L2 group_oper_assocA fix_var_val by simp
  ultimately show "?T = ?F" by (rule func_eq)
next
  let ?T = "LeftTranslation(G,P,g)"
  let ?F = "Fix1stVar(P,g)"
  from assms have "?T: G→G" and "?F: G→G"
    using group0_5_L1 group_oper_assocA fix_1st_var_fun by auto
  moreover from assms have "∀a∈G. ?T`(a) = ?F`(a)"
    using group0_5_L2 group_oper_assocA fix_var_val by simp
  ultimately show "?T = ?F" by (rule func_eq)
qed

text‹A lemma about translating sets.›

lemma (in group0) ltrans_image: assumes A1: "V⊆G" and A2: "x∈G"
  shows "LeftTranslation(G,P,x)``(V) = {x⋅v. v∈V}"
proof -
  from assms have "LeftTranslation(G,P,x)``(V) = {LeftTranslation(G,P,x)`(v). v∈V}"
    using group0_5_L1 func_imagedef by blast
  moreover from assms have "∀v∈V. LeftTranslation(G,P,x)`(v) = x⋅v"
    using group0_5_L2 by auto
  ultimately show ?thesis by auto
qed

text‹A technical lemma about solving equations with translations.›

lemma (in group0) ltrans_inv_in: assumes A1: "V⊆G" and A2: "y∈G" and
  A3: "x ∈ LeftTranslation(G,P,y)``(GroupInv(G,P)``(V))"
  shows "y ∈ LeftTranslation(G,P,x)``(V)"
proof -
  have "x∈G"
  proof -
    from A2 have "LeftTranslation(G,P,y):G→G" using group0_5_L1 by simp
    then have "LeftTranslation(G,P,y)``(GroupInv(G,P)``(V)) ⊆ G"
      using func1_1_L6 by simp
    with A3 show "x∈G" by auto
  qed
  have "∃v∈V. x = y⋅v¯"
  proof -
    have "GroupInv(G,P): G→G" using groupAssum group0_2_T2
      by simp
    with assms obtain z where "z ∈ GroupInv(G,P)``(V)" and "x = y⋅z"
      using func1_1_L6 ltrans_image by auto
    with A1 ‹GroupInv(G,P): G→G› show ?thesis using func_imagedef by auto
  qed
  then obtain v where "v∈V" and "x = y⋅v¯" by auto
  with A1 A2 have "y = x⋅v" using inv_cancel_two by auto
  with assms ‹x∈G› ‹v∈V› show ?thesis using ltrans_image by auto
qed

text‹We can look at the result of interval arithmetic operation as union of
  translated sets.›

lemma (in group0) image_ltrans_union: assumes "A⊆G" "B⊆G" shows
  "(P {lifted to subsets of} G)`⟨A,B⟩ = (⋃a∈A.  LeftTranslation(G,P,a)``(B))"
proof
  from assms have I: "(P {lifted to subsets of} G)`⟨A,B⟩ = {a⋅b . ⟨a,b⟩ ∈ A×B}"
    using group_oper_assocA lift_subsets_explained by simp
  { fix c assume "c ∈ (P {lifted to subsets of} G)`⟨A,B⟩"
    with I obtain a b where "c = a⋅b" and "a∈A"  "b∈B" by auto
    hence "c ∈ {a⋅b. b∈B}" by auto
    moreover from assms ‹a∈A› have 
      "LeftTranslation(G,P,a)``(B) = {a⋅b. b∈B}" using ltrans_image by auto
    ultimately have "c ∈ LeftTranslation(G,P,a)``(B)" by simp
    with ‹a∈A› have "c ∈ (⋃a∈A.  LeftTranslation(G,P,a)``(B))" by auto
  } thus "(P {lifted to subsets of} G)`⟨A,B⟩ ⊆ (⋃a∈A.  LeftTranslation(G,P,a)``(B))"
    by auto
  { fix c assume "c ∈ (⋃a∈A.  LeftTranslation(G,P,a)``(B))"
    then obtain a where "a∈A" and "c ∈ LeftTranslation(G,P,a)``(B)"
      by auto
    moreover from assms ‹a∈A› have "LeftTranslation(G,P,a)``(B) = {a⋅b. b∈B}"
      using ltrans_image by auto
    ultimately obtain b where "b∈B" and "c = a⋅b" by auto
    with I ‹a∈A› have "c ∈ (P {lifted to subsets of} G)`⟨A,B⟩" by auto
  } thus "(⋃a∈A. LeftTranslation(G,P,a)``(B)) ⊆ (P {lifted to subsets of} G)`⟨A,B⟩"
    by auto
qed

text‹If the neutral element belongs to a set, then an element of group belongs
  the translation of that set.›

lemma (in group0) neut_trans_elem: 
  assumes A1: "A⊆G" "g∈G" and A2: "𝟭∈A" 
  shows "g ∈ LeftTranslation(G,P,g)``(A)"
proof -
  from assms have "g⋅𝟭 ∈ LeftTranslation(G,P,g)``(A)"
    using ltrans_image by auto
  with A1 show ?thesis using group0_2_L2 by simp
qed

text‹The neutral element belongs to the translation of a set by the inverse
  of an element that belongs to it.›

lemma (in group0) elem_trans_neut: assumes A1: "A⊆G" and A2: "g∈A"
  shows "𝟭 ∈ LeftTranslation(G,P,g¯)``(A)"
proof -
  from assms have "g¯ ∈ G" using inverse_in_group by auto
  with assms have "g¯⋅g ∈ LeftTranslation(G,P,g¯)``(A)"
    using ltrans_image by auto
  moreover from assms have "g¯⋅g = 𝟭" using group0_2_L6 by auto
  ultimately show ?thesis by simp
qed

subsection‹Odd functions›

text‹This section is about odd functions.›

text‹Odd functions are those that commute with the group inverse:
  $f(a^{-1}) = (f(a))^{-1}.$›

definition
  "IsOdd(G,P,f) ≡ (∀a∈G. f`(GroupInv(G,P)`(a)) = GroupInv(G,P)`(f`(a)) )"

text‹Let's see the definition of an odd function in a more readable 
  notation.›

lemma (in group0) group0_6_L1: 
  shows "IsOdd(G,P,p) ⟷ ( ∀a∈G. p`(a¯) = (p`(a))¯ )"
  using IsOdd_def by simp

text‹We can express the definition of an odd function in two ways.›

lemma (in group0) group0_6_L2:
  assumes A1: "p : G→G" 
  shows 
  "(∀a∈G. p`(a¯) = (p`(a))¯) ⟷ (∀a∈G. (p`(a¯))¯ = p`(a))"
proof
  assume "∀a∈G. p`(a¯) = (p`(a))¯"
  with A1 show "∀a∈G. (p`(a¯))¯ = p`(a)"
    using apply_funtype group_inv_of_inv by simp
next assume A2: "∀a∈G. (p`(a¯))¯ = p`(a)"
  { fix a assume "a∈G"
    with A1 A2  have 
      "p`(a¯) ∈ G" and "((p`(a¯))¯)¯ =  (p`(a))¯"
    using apply_funtype inverse_in_group by auto
  then have "p`(a¯) = (p`(a))¯"
    using group_inv_of_inv by simp
  } then show "∀a∈G. p`(a¯) = (p`(a))¯" by simp
qed

end