theory test imports Main begin lemma test1: "((x::int) * y mod 2) mod 2 = (x * y) mod 2" by presburger lemma test2: "((x::int) + y mod 2) mod 2 = (x + y) mod 2" by presburger lemma zmod_eq_dvd_iff: "(x::int) mod n = y mod n \ n dvd x - y" proof assume H: "x mod n = y mod n" hence "x mod n - y mod n = 0" by simp hence "(x mod n - y mod n) mod n = 0" by simp hence "(x - y) mod n = 0" by (simp add: zmod_zdiff1_eq[symmetric]) thus "n dvd x - y" by (simp add: zdvd_iff_zmod_eq_0) next assume H: "n dvd x - y" then obtain k where k: "x-y = n*k" unfolding dvd_def by blast hence "x = n*k + y" by simp hence "x mod n = (n*k + y) mod n" by simp thus "x mod n = y mod n" by (simp add: zmod_zadd_left_eq) qed lemma int_mod_lemma: assumes xyn: "(x::int) mod n = y mod n" and xy:"y \ x" shows "\q. x = y + n * q" proof- from xy have th: "x - y = (x - y)" by presburger from xyn have "n dvd x - y" by (simp only: zmod_eq_dvd_iff[symmetric]) then show ?thesis using xy unfolding dvd_def apply clarsimp apply (rule_tac x="k" in exI) by arith qed lemma int_mod: "(x::int) mod n = y mod n \ (\q1 q2. x + n * q1 = y + n * q2)" (is "?lhs = ?rhs") proof assume H: "x mod n = y mod n" {assume xy: "x \ y" from H have th: "y mod n = x mod n" by simp from int_mod_lemma[OF th xy] have ?rhs apply clarify apply (rule_tac x="q" in exI) by (rule exI[where x="0"], simp)} moreover {assume xy: "y \ x" from int_mod_lemma[OF H xy] have ?rhs apply clarify apply (rule_tac x="0" in exI) by (rule_tac x="q" in exI, simp)} ultimately show ?rhs using linear[of x y] by blast next assume ?rhs then obtain q1 q2 where q12: "x + n * q1 = y + n * q2" by blast hence "(x + n * q1) mod n = (y + n * q2) mod n" by simp thus ?lhs by (simp add: zmod_zadd_right_eq) qed lemma test3: "((x::int) * y mod n) mod n = (x * y) mod n" unfolding zmod_simps mod_mod_trivial unfolding int_mod apply algebra done lemma test4: "((x::int) + y mod n) mod n = (x + y) mod n" unfolding zmod_simps mod_mod_trivial unfolding int_mod apply algebra done end