(* This file is a part of IsarMathLib - a library of formalized mathematics for Isabelle/Isar. Copyright (C) 2008 Slawomir Kolodynski This program is free software; Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met: 1. Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer. 2. Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution. 3. The name of the author may not be used to endorse or promote products derived from this software without specific prior written permission. THIS SOFTWARE IS PROVIDED BY THE AUTHOR ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. *) 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_fun 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_fun 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_fun 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‹Translation by neutral element does not move sets. › lemma (in group0) trans_neutral_image: assumes "V⊆G" shows "RightTranslation(G,P,𝟭)``(V) = V" and "LeftTranslation(G,P,𝟭)``(V) = V" using assms trans_neutral image_id_same by auto 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_fun fix_2nd_var_fun by auto moreover from assms have "∀a∈G. ?T`(a) = ?F`(a)" using group0_5_L2 group_oper_fun 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_fun fix_1st_var_fun by auto moreover from assms have "∀a∈G. ?T`(a) = ?F`(a)" using group0_5_L2 group_oper_fun fix_var_val by simp ultimately show "?T = ?F" by (rule func_eq) qed text‹A lemma demonstrating what is the left translation of a set› 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 lemma demonstrating what is the right translation of a set› lemma (in group0) rtrans_image: assumes A1: "V⊆G" and A2: "x∈G" shows "RightTranslation(G,P,x)``(V) = {v⋅x. v∈V}" proof - from assms have "RightTranslation(G,P,x)``(V) = {RightTranslation(G,P,x)`(v). v∈V}" using group0_5_L1 func_imagedef by blast moreover from assms have "∀v∈V. RightTranslation(G,P,x)`(v) = v⋅x" using group0_5_L2 by auto ultimately show ?thesis by auto qed text‹Right and left translations of a set are subsets of the group. Interestingly, we do not have to assume the set is a subset of the group. › lemma (in group0) lrtrans_in_group: assumes "x∈G" shows "LeftTranslation(G,P,x)``(V) ⊆ G" and "RightTranslation(G,P,x)``(V) ⊆ G" proof - from assms have "LeftTranslation(G,P,x):G→G" and "RightTranslation(G,P,x):G→G" using group0_5_L1 by auto then show "LeftTranslation(G,P,x)``(V) ⊆ G" and "RightTranslation(G,P,x)``(V) ⊆ G" using func1_1_L6(2) 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 left 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_fun 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‹ The right translation version of ‹image_ltrans_union› The proof follows the same schema.› lemma (in group0) image_rtrans_union: assumes "A⊆G" "B⊆G" shows "(P {lifted to subsets of} G)`⟨A,B⟩ = (⋃b∈B. RightTranslation(G,P,b)``(A))" proof from assms have I: "(P {lifted to subsets of} G)`⟨A,B⟩ = {a⋅b . ⟨a,b⟩ ∈ A×B}" using group_oper_fun 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. a∈A}" by auto moreover from assms ‹b∈B› have "RightTranslation(G,P,b)``(A) = {a⋅b. a∈A}" using rtrans_image by auto ultimately have "c ∈ RightTranslation(G,P,b)``(A)" by simp with ‹b∈B› have "c ∈ (⋃b∈B. RightTranslation(G,P,b)``(A))" by auto } thus "(P {lifted to subsets of} G)`⟨A,B⟩ ⊆ (⋃b∈B. RightTranslation(G,P,b)``(A))" by auto { fix c assume "c ∈ (⋃b∈B. RightTranslation(G,P,b)``(A))" then obtain b where "b∈B" and "c ∈ RightTranslation(G,P,b)``(A)" by auto moreover from assms ‹b∈B› have "RightTranslation(G,P,b)``(A) = {a⋅b. a∈A}" using rtrans_image by auto ultimately obtain a where "a∈A" and "c = a⋅b" by auto with I ‹b∈B› have "c ∈ (P {lifted to subsets of} G)`⟨A,B⟩" by auto } thus "(⋃b∈B. RightTranslation(G,P,b)``(A)) ⊆ (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)" "g ∈ RightTranslation(G,P,g)``(A)" proof - from assms have "g⋅𝟭 ∈ LeftTranslation(G,P,g)``(A)" using ltrans_image by auto with A1 show "g ∈ LeftTranslation(G,P,g)``(A)" using group0_2_L2 by simp from assms have "𝟭⋅g ∈ RightTranslation(G,P,g)``(A)" using rtrans_image by auto with A1 show "g ∈ RightTranslation(G,P,g)``(A)" 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)" "𝟭 ∈ RightTranslation(G,P,g¯)``(A)" proof - from assms have ginv:"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 "𝟭 ∈ LeftTranslation(G,P,g¯)``(A)" by simp from ginv assms have "g⋅g¯ ∈ RightTranslation(G,P,g¯)``(A)" using rtrans_image by auto moreover from assms have "g⋅g¯ = 𝟭" using group0_2_L6 by auto ultimately show "𝟭 ∈ RightTranslation(G,P,g¯)``(A)" 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 subsection‹Subgroups and interval arithmetic› text‹ The section ‹Binary operations› in the ‹func_ZF› theory defines the notion of "lifting operation to subsets". In short, every binary operation $f:X\times X \longrightarrow X $ on a set $X$ defines an operation on the subsets of $X$ defined by $F(A,B) = \{ f\langle x,y \rangle | x\in A, y\in B\}$. In the group context using multiplicative notation we can write this as $H\cdot K = \{ x\cdot y | x\in A, y\in B\}$. Similarly we can define $H^{-1}=\{ x^{-1} | x\in H\}$. In this section we study properties of these derived operation and how they relate to the concept of subgroups.› text‹The next locale extends the ‹groups0› locale with notation related to interval arithmetics.› locale group4 = group0 + fixes sdot (infixl "\<sdot>" 70) defines sdot_def [simp]: "A\<sdot>B ≡ (P {lifted to subsets of} G)`⟨A,B⟩" fixes sinv ("_\<sinv> " [90] 91) defines sinv_def[simp]: "A\<sinv> ≡ GroupInv(G,P)``(A)" text‹The next lemma shows a somewhat more explicit way of defining the product of two subsets of a group.› lemma (in group4) interval_prod: assumes "A⊆G" "B⊆G" shows "A\<sdot>B = {x⋅y. ⟨x,y⟩ ∈ A×B}" using assms group_oper_fun lift_subsets_explained by auto text‹Product of elements of subsets of the group is in the set product of those subsets› lemma (in group4) interval_prod_el: assumes "A⊆G" "B⊆G" "x∈A" "y∈B" shows "x⋅y ∈ A\<sdot>B" using assms interval_prod by auto text‹An alternative definition of a group inverse of a set.› lemma (in group4) interval_inv: assumes "A⊆G" shows "A\<sinv> = {x¯.x∈A}" proof - from groupAssum have "GroupInv(G,P):G→G" using group0_2_T2 by simp with assms show "A\<sinv> = {x¯.x∈A}" using func_imagedef by simp qed text‹Group inverse of a set is a subset of the group. Interestingly we don't need to assume the set is a subset of the group.› lemma (in group4) interval_inv_cl: shows "A\<sinv> ⊆ G" proof - from groupAssum have "GroupInv(G,P):G→G" using group0_2_T2 by simp then show "A\<sinv> ⊆ G" using func1_1_L6(2) by simp qed text‹The product of two subsets of a group is a subset of the group.› lemma (in group4) interval_prod_closed: assumes "A⊆G" "B⊆G" shows "A\<sdot>B ⊆ G" proof fix z assume "z ∈ A\<sdot>B" with assms obtain x y where "x∈A" "y∈B" "z=x⋅y" using interval_prod by auto with assms show "z∈G" using group_op_closed by auto qed text‹ The product of sets operation is associative.› lemma (in group4) interval_prod_assoc: assumes "A⊆G" "B⊆G" "C⊆G" shows "A\<sdot>B\<sdot>C = A\<sdot>(B\<sdot>C)" proof - from groupAssum have "(P {lifted to subsets of} G) {is associative on} Pow(G)" unfolding IsAgroup_def IsAmonoid_def using lift_subset_assoc by simp with assms show ?thesis unfolding IsAssociative_def by auto qed text‹ A simple rearrangement following from associativity of the product of sets operation.› lemma (in group4) interval_prod_rearr1: assumes "A⊆G" "B⊆G" "C⊆G" "D⊆G" shows "A\<sdot>B\<sdot>(C\<sdot>D) = A\<sdot>(B\<sdot>C)\<sdot>D" proof - from assms(1,2) have "A\<sdot>B ⊆ G" using interval_prod_closed by simp with assms(3,4) have "A\<sdot>B\<sdot>(C\<sdot>D) = A\<sdot>B\<sdot>C\<sdot>D" using interval_prod_assoc by simp also from assms(1,2,3) have "A\<sdot>B\<sdot>C\<sdot>D = A\<sdot>(B\<sdot>C)\<sdot>D" using interval_prod_assoc by simp finally show ?thesis by simp qed text‹A subset $A$ of the group is closed with respect to the group operation iff $A\cdot A \subseteq A$. › lemma (in group4) subset_gr_op_cl: assumes "A⊆G" shows "(A {is closed under} P) ⟷ A\<sdot>A ⊆ A" proof assume "A {is closed under} P" { fix z assume "z ∈ A\<sdot>A" with assms obtain x y where "x∈A" "y∈A" and "z=x⋅y" using interval_prod by auto with ‹A {is closed under} P› have "z∈A" unfolding IsOpClosed_def by simp } thus "A\<sdot>A ⊆ A" by auto next assume "A\<sdot>A ⊆ A" { fix x y assume "x∈A" "y∈A" with assms have "x⋅y ∈ A\<sdot>A" using interval_prod by auto with ‹A\<sdot>A ⊆ A› have "x⋅y ∈ A" by auto } then show "A {is closed under} P" unfolding IsOpClosed_def by simp qed text‹Inverse and square of a subgroup is this subgroup.› lemma (in group4) subgroup_inv_sq: assumes "IsAsubgroup(H,P)" shows "H\<sinv> = H" and "H\<sdot>H = H" proof from assms have "H⊆G" using group0_3_L2 by simp with assms show "H\<sinv> ⊆ H" using interval_inv group0_3_T3A by auto { fix x assume "x∈H" with assms have "(x¯)¯ ∈ {y¯.y∈H}" using group0_3_T3A by auto moreover from ‹x∈H› ‹H⊆G› have "(x¯)¯ = x" using group_inv_of_inv by auto ultimately have "x ∈ {y¯.y∈H}" by auto with ‹H⊆G› have "x ∈ H\<sinv>" using interval_inv by simp } thus "H ⊆ H\<sinv>" by auto from assms have "H {is closed under} P" using group0_3_L6 unfolding IsOpClosed_def by simp with assms have "H\<sdot>H ⊆ H" using subset_gr_op_cl group0_3_L2 by simp moreover { fix x assume "x∈H" with assms have "x∈G" using group0_3_L2 by auto from assms ‹H⊆G› ‹x∈H› have "x⋅𝟭 ∈ H\<sdot>H" using group0_3_L5 interval_prod by auto with ‹x∈G› have "x ∈ H\<sdot>H" using group0_2_L2 by simp } hence "H ⊆ H\<sdot>H" by auto ultimately show "H\<sdot>H = H" by auto qed text‹Inverse of a product two sets is a product of inverses with the reversed order.› lemma (in group4) interval_prod_inv: assumes "A⊆G" "B⊆G" shows "(A\<sdot>B)\<sinv> = {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" "(A\<sdot>B)\<sinv> = {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" "(A\<sdot>B)\<sinv> = (B\<sinv>)\<sdot>(A\<sinv>)" proof - from assms have "(A\<sdot>B) ⊆ G" using interval_prod_closed by simp then have I: "(A\<sdot>B)\<sinv> = {z¯.z∈A\<sdot>B}" using interval_inv by simp show II: "(A\<sdot>B)\<sinv> = {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" proof { fix p assume "p ∈ (A\<sdot>B)\<sinv>" with I obtain z where "p=z¯" and "z∈A\<sdot>B" by auto with assms obtain x y where "⟨x,y⟩ ∈ A×B" and "z=x⋅y" using interval_prod by auto with ‹p=z¯› have "p∈{(x⋅y)¯.⟨x,y⟩ ∈ A×B}" by auto } thus "(A\<sdot>B)\<sinv> ⊆ {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" by blast { fix p assume "p∈{(x⋅y)¯.⟨x,y⟩ ∈ A×B}" then obtain x y where "x∈A" "y∈B" and "p=(x⋅y)¯" by auto with assms ‹(A\<sdot>B) ⊆ G› have "p∈(A\<sdot>B)\<sinv>" using interval_prod interval_inv by auto } thus "{(x⋅y)¯.⟨x,y⟩ ∈ A×B} ⊆ (A\<sdot>B)\<sinv>" by blast qed have "{(x⋅y)¯.⟨x,y⟩ ∈ A×B} = {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" proof { fix p assume "p ∈ {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" then obtain x y where "x∈A" "y∈B" and "p=(x⋅y)¯" by auto with assms have "y¯⋅x¯ = (x⋅y)¯" using group_inv_of_two by auto with ‹p=(x⋅y)¯› have "p = y¯⋅x¯" by simp with ‹x∈A› ‹y∈B› have "p∈{y¯⋅x¯.⟨x,y⟩ ∈ A×B}" by auto } thus "{(x⋅y)¯.⟨x,y⟩ ∈ A×B} ⊆ {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" by blast { fix p assume "p∈{y¯⋅x¯.⟨x,y⟩ ∈ A×B}" then obtain x y where "x∈A" "y∈B" and "p=y¯⋅x¯" by auto with assms have "p = (x⋅y)¯" using group_inv_of_two by auto with ‹x∈A› ‹y∈B› have "p ∈ {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" by auto } thus "{y¯⋅x¯.⟨x,y⟩ ∈ A×B} ⊆ {(x⋅y)¯.⟨x,y⟩ ∈ A×B}" by blast qed with II show III: "(A\<sdot>B)\<sinv> = {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" by simp have "{y¯⋅x¯.⟨x,y⟩ ∈ A×B} = (B\<sinv>)\<sdot>(A\<sinv>)" proof { fix p assume "p∈{y¯⋅x¯.⟨x,y⟩ ∈ A×B}" then obtain x y where "x∈A" "y∈B" and "p=y¯⋅x¯" by auto with assms have "y¯ ∈ (B\<sinv>)" and "x¯ ∈ (A\<sinv>)" using interval_inv by auto with ‹p=y¯⋅x¯› have "p ∈ (B\<sinv>)\<sdot>(A\<sinv>)" using interval_inv_cl interval_prod by auto } thus "{y¯⋅x¯.⟨x,y⟩ ∈ A×B} ⊆ (B\<sinv>)\<sdot>(A\<sinv>)" by blast { fix p assume "p ∈ (B\<sinv>)\<sdot>(A\<sinv>)" then obtain y x where "y∈B\<sinv>" "x∈A\<sinv>" and "p=y⋅x" using interval_inv_cl interval_prod by auto with assms obtain x⇩_{1}y⇩_{1}where "x⇩_{1}∈ A" "y⇩_{1}∈ B" and "x=x⇩_{1}¯" "y=y⇩_{1}¯" using interval_inv by auto with ‹p=y⋅x› have "p ∈ {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" by auto } thus "(B\<sinv>)\<sdot>(A\<sinv>) ⊆ {y¯⋅x¯.⟨x,y⟩ ∈ A×B}" by blast qed with III show "(A\<sdot>B)\<sinv> = (B\<sinv>)\<sdot>(A\<sinv>)" by simp qed text‹ If $H,K$ are subgroups then $H\cdot K$ is a subgroup iff $H\cdot K = K\cdot H$. › theorem (in group4) prod_subgr_subgr: assumes "IsAsubgroup(H,P)" and "IsAsubgroup(K,P)" shows "IsAsubgroup(H\<sdot>K,P) ⟷ H\<sdot>K = K\<sdot>H" proof assume "IsAsubgroup(H\<sdot>K,P)" then have "(H\<sdot>K)\<sinv> = H\<sdot>K" using subgroup_inv_sq(1) by simp with assms show "H\<sdot>K = K\<sdot>H" using group0_3_L2 interval_prod_inv subgroup_inv_sq(1) by auto next from assms have "H⊆G" and "K⊆G" using group0_3_L2 by auto have I: "H\<sdot>K ≠ 0" proof - let ?x = "𝟭" let ?y = "𝟭" from assms have "?x⋅?y ∈ (H\<sdot>K)" using group0_3_L5 group0_3_L2 interval_prod by auto thus ?thesis by auto qed from ‹H⊆G› ‹K⊆G› have II: "H\<sdot>K ⊆ G" using interval_prod_closed by simp assume "H\<sdot>K = K\<sdot>H" have III: "(H\<sdot>K){is closed under} P" proof - have "(H\<sdot>K)\<sdot>(H\<sdot>K) = H\<sdot>K" proof - from ‹H⊆G› ‹K⊆G› have "(H\<sdot>K)\<sdot>(H\<sdot>K) = H\<sdot>(K\<sdot>H)\<sdot>K" using interval_prod_rearr1 by simp also from ‹H\<sdot>K = K\<sdot>H› have "... = H\<sdot>(H\<sdot>K)\<sdot>K" by simp also from ‹H⊆G› ‹K⊆G› have "... = (H\<sdot>H)\<sdot>(K\<sdot>K)" using interval_prod_rearr1 by simp also from assms have "... = H\<sdot>K" using subgroup_inv_sq(2) by simp finally show ?thesis by simp qed with ‹H\<sdot>K ⊆ G› show ?thesis using subset_gr_op_cl by simp qed have IV: "∀x ∈ H\<sdot>K. x¯ ∈ H\<sdot>K" proof - { fix x assume "x ∈ H\<sdot>K" with ‹H\<sdot>K ⊆ G› have "x¯ ∈ (H\<sdot>K)\<sinv>" using interval_inv by auto with assms ‹H⊆G› ‹K⊆G› ‹H\<sdot>K = K\<sdot>H› have "x¯ ∈ H\<sdot>K" using interval_prod_inv subgroup_inv_sq(1) by simp } thus ?thesis by auto qed from I II III IV show "IsAsubgroup(H\<sdot>K,P)" using group0_3_T3 by simp qed end