abbreviation Fn_project_abbr (infix‹↓Fn›55) where"f ↓Fn v ≡ Fn_project⋅f⋅v"
lemma [simp]: "⊥↓Fn x = ⊥" "(B⋅b) ↓Fn x = ⊥" by (fixrec_simp)+
fixrec B_project :: "Value → Value → Value → Value"where "B_project⋅(B⋅db)⋅v1⋅v2 = (if undiscr db then v1 else v2)"
lemma [simp]: "B_project⋅(B⋅(Discr b))⋅v1⋅v2 = (if b then v1 else v2)" "B_project⋅⊥⋅v1⋅v2 = ⊥" "B_project⋅(Fn⋅f)⋅v1⋅v2 = ⊥" by fixrec_simp+
text‹
chain in the domain @{typ Value} is either always bottom, or eventually @{term "Fn"} of another
›
lemma Value_chainE[consumes 1, case_names bot B Fn]: assumes"chain Y" obtains"Y = (λ _ . ⊥)" |
n b where"Y = (λ m. (if m < n then ⊥ else B⋅b))" |
n Y' where"Y = (λ m. (if m < n then ⊥ else Fn⋅(Y' (m-n))))""chain Y'" proof(cases "Y = (λ _ . ⊥)") case True thus ?thesis by (rule that(1)) next case False hence"∃ i. Y i ≠⊥"by auto hence"∃ n. Y n ≠⊥∧ (∀m. Y m ≠⊥⟶ m ≥ n)" by (rule exE)(rule ex_has_least_nat) thenobtain n where"Y n ≠⊥"and"∀m. m < n ⟶ Y m = ⊥"by fastforce hence"(∃ f. Y n = Fn⋅f) ∨ (∃ b. Y n = B⋅b)"by (metis Value.exhaust) thus ?thesis proof assume"(∃ f. Y n = Fn⋅f)" thenobtain f where"Y n = Fn ⋅ f"by blast
{ fix i from‹chain Y›have"Y n ⊑ Y (i+n)"by (metis chain_mono le_add2) with‹Y n = _› have"∃ g. (Y (i+n) = Fn ⋅ g)" by (metis Value.dist_les(1) Value.exhaust below_bottom_iff)
} thenobtain Y' where Y': "∧ i. Y (i + n) = Fn ⋅ (Y' i)"by metis
have"Y = (λm. if m < n then ⊥ else Fn⋅(Y' (m - n)))" using‹∀m. _› Y' by (metis add_diff_inverse add.commute) moreover have"chain Y'"using‹chain Y› by (auto intro!:chainI elim: chainE simp add: Value.inverts[symmetric] Y'[symmetric] simp del: Value.inverts) ultimately show ?thesis by (rule that(3)) next assume"(∃ b. Y n = B⋅b)" thenobtain b where"Y n = B⋅b"by blast
{ fix i from‹chain Y›have"Y n ⊑ Y (i+n)"by (metis chain_mono le_add2) with‹Y n = _› have"Y (i+n) = B⋅b" by (metis Value.dist_les(2) Value.exhaust Value.inverts(2) below_bottom_iff discrete_cpo)
} hence Y': "∧ i. Y (i + n) = B⋅b"by metis
have"Y = (λm. if m < n then ⊥ else B⋅b)" using‹∀m. _› Y' by (metis add_diff_inverse add.commute) thus ?thesis by (rule that(2)) qed qed
Die Informationen auf dieser Webseite wurden
nach bestem Wissen sorgfältig zusammengestellt. Es wird jedoch weder Vollständigkeit, noch Richtigkeit,
noch Qualität der bereit gestellten Informationen zugesichert.
Bemerkung:
Die farbliche Syntaxdarstellung und die Messung sind noch experimentell.