What kind of functions preserve properties of closure?

I'm trying to prove the following lemmas:

lemma tranclp_fun_preserve:
  "(⋀x y. x ≠ y ⟹ f x ≠ f y) ⟹
   (⋀x y. f x ≠ f y ⟹ x ≠ y) ⟹
   (⋀x y. f x = f y ⟹ x = y) ⟹
   (λ x y. P x y)⇧+⇧+ (f x) (f y) ⟹ (λ x y. P (f x) (f y))⇧+⇧+ x y"
  apply (erule tranclp.cases)
  apply blast

lemma tranclp_fun_preserve2:
  "(⋀x y. x ≠ y ⟹ f x ≠ f y) ⟹
   (⋀x y. f x ≠ f y ⟹ x ≠ y) ⟹
   (⋀x y. f x = f y ⟹ x = y) ⟹
   (λ x y. P (f x) (f y))⇧+⇧+ x y ⟹ (λ x y. P x y)⇧+⇧+ (f x) (f y)"
  apply (erule tranclp.cases)
  apply blast

However, I'm stuck. Should I choose another set of assumptions for the function f? Could you suggest how to prove the lemmas tranclp_fun_preserve and tranclp_fun_preserve2?

UPDATE

My function is injective with a special property described at the end. I'm afraid that the following example is too long. However, I want to make it a little bit more realistic. Here are two auxiliary types errorable and nullable:

(*** errorable ***)

notation
  bot ("⊥")

typedef 'a errorable ("_⇩⊥" [21] 21) = "UNIV :: 'a option set" ..

definition errorable :: "'a ⇒ 'a errorable" ("_⇩⊥" [1000] 1000) where
  "errorable x = Abs_errorable (Some x)"

instantiation errorable :: (type) bot
begin
definition "⊥ ≡ Abs_errorable None"
instance ..
end

free_constructors case_errorable for
  errorable
| "⊥ :: 'a errorable"
  apply (metis Rep_errorable_inverse bot_errorable_def errorable_def not_Some_eq)
  apply (metis Abs_errorable_inverse UNIV_I errorable_def option.inject)
  by (simp add: Abs_errorable_inject bot_errorable_def errorable_def)

(*** nullable ***)

class opt =
  fixes null :: "'a" ("ε")

typedef 'a nullable ("_⇩□" [21] 21) = "UNIV :: 'a option set" ..

definition nullable :: "'a ⇒ 'a nullable" ("_⇩□" [1000] 1000) where
  "nullable x = Abs_nullable (Some x)"

instantiation nullable :: (type) opt
begin
definition "ε ≡ Abs_nullable None"
instance ..
end

free_constructors case_nullable for
  nullable
| "ε :: 'a nullable"
  apply (metis Rep_nullable_inverse null_nullable_def nullable_def option.collapse)
  apply (simp add: Abs_nullable_inject nullable_def)
  by (metis Abs_nullable_inverse UNIV_I null_nullable_def nullable_def option.distinct(1))

Two kinds of values:

datatype any = BoolVal "bool⇩⊥" | NatVal "nat⇩⊥" | RealVal "real⇩⊥" | InvalidAny unit

datatype oany = OBoolVal "bool⇩⊥⇩□" | ONatVal "nat⇩⊥⇩□" | ORealVal "real⇩⊥⇩□" | OInvalidAny "unit⇩□"

Here is a concrete example of the function f (any_to_oany), it's injective:

inductive any_oany :: "any ⇒ oany ⇒ bool" where
  "any_oany (BoolVal x) (OBoolVal x⇩□)"
| "any_oany (NatVal x) (ONatVal x⇩□)"
| "any_oany (RealVal x) (ORealVal x⇩□)"
| "any_oany (InvalidAny x) (OInvalidAny x⇩□)"

fun any_to_oany :: "any ⇒ oany" where
  "any_to_oany (BoolVal x) = (OBoolVal x⇩□)"
| "any_to_oany (NatVal x) = (ONatVal x⇩□)"
| "any_to_oany (RealVal x) = (ORealVal x⇩□)"
| "any_to_oany (InvalidAny x) = (OInvalidAny x⇩□)"

lemma any_oany_eq_fun:
  "any_oany x y ⟷ any_to_oany x = y"
  by (cases x; cases y; auto simp: any_oany.simps)

Here is a concrete example of the relation P (cast_oany):

inductive cast_any :: "any ⇒ any ⇒ bool" where
  "cast_any (BoolVal ⊥) (InvalidAny ())"
| "cast_any (NatVal ⊥) (RealVal ⊥)"
| "cast_any (NatVal x⇩⊥) (RealVal (real x)⇩⊥)"
| "cast_any (RealVal ⊥) (InvalidAny ())"

inductive cast_oany :: "oany ⇒ oany ⇒ bool" where
  "cast_any x y ⟹ any_oany x ox ⟹ any_oany y oy ⟹
   cast_oany ox oy"
| "cast_oany (OBoolVal ε) (OInvalidAny ε)"
| "cast_oany (ONatVal ε) (ORealVal ε)"
| "cast_oany (ORealVal ε) (OInvalidAny ε)"

I proved equivalence of cast_any and cast_oany on any:

lemma cast_any_implies_cast_oany:
  "cast_any x y ⟹ cast_oany (any_to_oany x) (any_to_oany y)"
  by (simp add: any_oany_eq_fun cast_oany.intros(1))

lemma cast_oany_implies_cast_any:
  "cast_oany (any_to_oany x) (any_to_oany y) ⟹ cast_any x y"
  by (cases x; cases y; simp add: any_oany.simps cast_oany.simps)

And my final goal is to prove similar lemmas for the transitive closures of these relations:

lemma trancl_cast_oany_implies_cast_any:
  "cast_oany⇧+⇧+ (any_to_oany x) (any_to_oany y) ⟹ cast_any⇧+⇧+ x y"

lemma trancl_cast_any_implies_cast_oany:
  "cast_any⇧+⇧+ x y ⟹ cast_oany⇧+⇧+ (any_to_oany x) (any_to_oany y)"

I proved the following intermediate lemmas:

lemma trancl_cast_oany_implies_cast_any':
  "(λx y. cast_oany (any_to_oany x) (any_to_oany y))⇧+⇧+ x y ⟹
   cast_any⇧+⇧+ x y"
  apply (erule tranclp_trans_induct)
  apply (simp add: cast_oany_implies_cast_any tranclp.r_into_trancl)
  by auto

lemma trancl_cast_any_implies_cast_oany':
  "cast_any⇧+⇧+ x y ⟹
   (λx y. cast_oany (any_to_oany x) (any_to_oany y))⇧+⇧+ x y"
  apply (erule tranclp_trans_induct)
  apply (simp add: cast_any_implies_cast_oany tranclp.r_into_trancl)
  by auto

At last, if I could prove the following lemmas from the original question, then I will able to prove my goal lemmas.

lemma tranclp_fun_preserve:
  "cast_oany⇧+⇧+ (any_to_oany x) (any_to_oany y) ⟹
   (λx y. cast_oany (any_to_oany x) (any_to_oany y))⇧+⇧+ x y"

lemma tranclp_fun_preserve2:
  "(λx y. cast_oany (any_to_oany x) (any_to_oany y))⇧+⇧+ x y ⟹
   cast_oany⇧+⇧+ (any_to_oany x) (any_to_oany y)"

In this paragraph I provide an important property of the function any_to_oany. The set of oany values consists of two parts:

nulls (OBoolVal ε, ONatVal ε, ORealVal ε, OInvalidAny ε)
all other values.

The relation cast_oany relates the values inside the first part and inside the second part separately. There is no relation between the values from different parts. The function any_to_oany maps values only to the second part. I don't know what is the right name of this property: subsets 1 and 2 are "closed" with respect to the relation cast_oany. And the function any_to_oany maps values only to one subset, and it's bijective on this subset.

Solution

The answer presented below is a substantial revision of the original answer. The original answer is available through the revision history.

Effectively, in the course of the initial revisions it was established that the question comes down to merely proving that bijective functions between two sets preserve the properties of closure. The solution below presents the relevant proofs without the application-specific context (the answer also combines some of the amendments to the original answer that were made by the author of the thread):

section ‹Extension of the theory @{text Transitive_Closure}›
theory Transitive_Closure_Ext
  imports 
    Complex_Main
    "HOL-Library.FuncSet"
begin

lemma trancl_subset_trancl: "r ⊆ s⇧+ ⟹ r⇧+ ⊆ s⇧+"
  by (metis subsetI trancl_id trancl_mono trans_trancl)

lemma mono_tranclp[mono]: "(⋀a b. R a b ⟶ S a b) ⟹ R⇧+⇧+ a b ⟶ S⇧+⇧+ a b"
  apply(rule) using trancl_mono[to_pred] by blast

lemma tranclp_mono: "R ≤ S ⟹ R⇧+⇧+ ≤ S⇧+⇧+"
  by (metis (full_types) mono_tranclp predicate2D predicate2I)

lemma preserve_tranclp:
  assumes "⋀x y. R x y ⟹ S (f x) (f y)"
  shows "R⇧+⇧+ x y ⟹ S⇧+⇧+ (f x) (f y)"
proof -
  assume Rpp: "R⇧+⇧+ x y"
  define P where P: "P = (λx y. S⇧+⇧+ (f x) (f y))" 
  define r where r: "r = (λx y. S (f x) (f y))"
  have major: "r⇧+⇧+ x y" 
    by (insert assms Rpp r; erule tranclp_trans_induct; auto)
  have cases_1: "r x y ⟹ P x y" for x y unfolding r P by simp
  have cases_2: "r⇧+⇧+ x y ⟹ P x y ⟹ r⇧+⇧+ y z ⟹ P y z ⟹ P x z" for x y z
    unfolding P by simp
  from major cases_1 cases_2 have "P x y" by (rule tranclp_trans_induct)
  thus "S⇧+⇧+ (f x) (f y)" unfolding P . 
qed

lemma preserve_trancl:
  assumes "map_prod f f ` r ⊆ s"
  shows "map_prod f f ` r⇧+ ⊆ s⇧+"
proof -
  from assms have "(x, y) ∈ r ⟶ (f x, f y) ∈ s" for x y by auto
  then have "(x, y) ∈ r⇧+ ⟶ (f x, f y) ∈ s⇧+" for x y
  by (metis preserve_tranclp[to_set])
  thus "map_prod f f ` r⇧+ ⊆ s⇧+" by clarsimp
qed

lemma preserve_tranclp_inv:
  assumes bij_f: "bij_betw f a b"
    and R: "⋀x y. R x y ⟹ (x, y) ∈ a × a"
    and S: "⋀x y. S x y ⟹ (x, y) ∈ b × b"
    and S_R: "⋀x y. (x, y) ∈ a × a ⟹ S (f x) (f y) ⟹ R x y"
  shows "(x, y) ∈ a × a ⟹ S⇧+⇧+ (f x) (f y) ⟹ R⇧+⇧+ x y"
proof -
  assume x_y_in_aa: "(x, y) ∈ a × a"
  assume Spp: "S⇧+⇧+ (f x) (f y)"
  define g where g: "g = the_inv_into a f"
  define gr where gr: "gr = restrict g b"
  define P where P: "P = (λx y. (x, y) ∈ b × b ⟶ R⇧+⇧+ (gr x) (gr y))"
  from Spp have fx_fy_in_bb: "(f x, f y) ∈ b × b"  
    using S by (metis converse_tranclpE mem_Sigma_iff tranclp.cases)
  have cases_1: "S x y ⟹ P x y" for x y unfolding P
  proof(rule impI)
    assume Sxy: "S x y" and xy_in_bb: "(x, y) ∈ b × b"
    with bij_f have gr_in_aa: "(gr x, gr y) ∈ a × a" 
      unfolding gr g apply(auto)
      using bij_betwE bij_betw_the_inv_into by blast+
    from bij_f xy_in_bb have "x = f (gr x)" and "y = f (gr y)"
      unfolding gr g using f_the_inv_into_f_bij_betw by fastforce+
    with Sxy have S_fgrx_fgry: "S (f (gr x)) (f (gr y))" by simp
    from gr_in_aa S_fgrx_fgry have "R (gr x) (gr y)" by (rule S_R)
    thus "R⇧+⇧+ (gr x) (gr y)" ..
  qed
  with bij_f S have 
    "S⇧+⇧+ x y ⟹ S⇧+⇧+ y z ⟹ x ∈ b ⟹ z ∈ b ⟹ y ∈ b" for x y z
    by (auto dest: SigmaD1 tranclpD)
  with P have cases_2: 
    "S⇧+⇧+ x y ⟹ P x y ⟹ S⇧+⇧+ y z ⟹ P y z ⟹ P x z" for x y z
    by auto
  from Spp cases_1 cases_2 have "P (f x) (f y)" by (rule tranclp_trans_induct)
  with bij_f fx_fy_in_bb x_y_in_aa show "R⇧+⇧+ x y" 
    unfolding P gr g restrict_def bij_betw_def by (simp add: the_inv_into_f_f)
qed

lemma preserve_trancl_inv:
  assumes bij_f: "bij_betw f a b"
    and r_in_aa: "r ⊆ a × a"
    and s_in_bb: "s ⊆ b × b"
    and s_r: "(map_prod f f -` s) ∩ (a × a) ⊆ r ∩ (a × a)"
  shows "(map_prod f f -` s⇧+) ∩ (a × a) ⊆ r⇧+ ∩ (a × a)"
proof -
  from r_in_aa have r_in_aa_set: 
    "(x, y) ∈ r ⟹ (x, y) ∈ a × a" for x y by auto
  from s_in_bb have s_in_bb_set: "⋀x y. (x, y) ∈ s ⟹ (x, y) ∈ b × b" by auto
  from s_r have s_r_set: 
    "(x, y) ∈ a × a ⟹ (f x, f y) ∈ s ⟹ (x, y) ∈ r" for x y
    unfolding map_prod_def by auto
  from bij_f r_in_aa_set s_in_bb_set s_r_set have 
    "(x, y) ∈ a × a ⟹ (f x, f y) ∈ s⇧+ ⟹ (x, y) ∈ r⇧+" for x y
    by (rule preserve_tranclp_inv[to_set])
  thus ?thesis unfolding map_prod_def by clarsimp
qed

lemma preserve_rtranclp:
  assumes "⋀x y. R x y ⟹ S (f x) (f y)"
  shows "R⇧*⇧* x y ⟹ S⇧*⇧* (f x) (f y)"
  by (insert assms, metis Nitpick.rtranclp_unfold preserve_tranclp)

lemma preserve_rtrancl:
  assumes "map_prod f f ` r ⊆ s"
  shows "map_prod f f ` r⇧* ⊆ s⇧*"
proof -
  from assms have "(x, y) ∈ r ⟶ (f x, f y) ∈ s" for x y by auto
  then have "(x, y) ∈ r⇧* ⟶ (f x, f y) ∈ s⇧*" for x y 
    by (metis preserve_rtranclp[to_set])
  thus "map_prod f f ` r⇧* ⊆ s⇧*" by clarsimp
qed

lemma preserve_rtranclp_inv:
  assumes bij_f: "bij_betw f a b"
    and "⋀x y. R x y ⟹ (x, y) ∈ a × a"
    and "⋀x y. S x y ⟹ (x, y) ∈ b × b"
    and "⋀x y. (x, y) ∈ a × a ⟹ S (f x) (f y) ⟹ R x y"
  shows "(x, y) ∈ a × a ⟹ S⇧*⇧* (f x) (f y) ⟹ R⇧*⇧* x y"
proof -
  assume xy_in_aa: "(x, y) ∈ a × a" and Spp: "S⇧*⇧* (f x) (f y)" 
  show "R⇧*⇧* x y"
  proof(cases "f x ≠ f y")
    case True show ?thesis
    proof -
      from True Spp obtain z where "S⇧*⇧* (f x) z" and "S z (f y)" 
        by (auto elim: rtranclp.cases)
      then have "S⇧+⇧+ (f x) (f y)" by (rule rtranclp_into_tranclp1)
      with assms xy_in_aa have "R⇧+⇧+ x y" by (rule preserve_tranclp_inv)
      thus ?thesis by simp
    qed
  next
    case False show ?thesis
    proof -
      from False xy_in_aa bij_f have "x = y" 
        unfolding bij_betw_def by (auto dest: SigmaD1 SigmaD2 inj_onD)
      thus "R⇧*⇧* x y" by simp 
    qed
  qed
qed

lemma preserve_rtrancl_inv:
  assumes bij_f: "bij_betw f a b"
    and r_in_aa: "r ⊆ a × a"
    and s_in_bb: "s ⊆ b × b"
    and as_s_r: "(map_prod f f -` s) ∩ (a × a) ⊆ r ∩ (a × a)"
  shows "(map_prod f f -` s⇧*) ∩ (a × a) ⊆ r⇧* ∩ (a × a)"
proof -
  from r_in_aa have r_in_aa_set: 
    "(x, y) ∈ r ⟹ (x, y) ∈ a × a" for x y by auto
  from s_in_bb have s_in_bb_set: 
    "(x, y) ∈ s ⟹ (x, y) ∈ b × b" for x y by auto
  from as_s_r have s_r: 
    "(x, y) ∈ a × a ⟹ (f x, f y) ∈ s ⟹ (x, y) ∈ r" for x y 
    unfolding map_prod_def by auto
  from bij_f r_in_aa_set s_in_bb_set s_r have 
    "(x, y) ∈ a × a ⟹ (f x, f y) ∈ s⇧* ⟹ (x, y) ∈ r⇧*" for x y
    by (rule preserve_rtranclp_inv[to_set])
  thus ?thesis unfolding map_prod_def by clarsimp
qed

end