Theorem moddiffl

Description: Value of the modulo operation rewritten to give two ways of expressing the quotient when " A is divided by B using Euclidean division." Multiplying both sides by B , this implies that A mod B differs from A by an integer multiple of B . (Contributed by Jeff Madsen, 17-Jun-2010) (Revised by Mario Carneiro, 6-Sep-2016)

	Ref	Expression
Assertion	moddiffl	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 − ( 𝐴 mod 𝐵 ) ) / 𝐵 ) = ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) )

Proof

Step	Hyp	Ref	Expression
1		modval	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 mod 𝐵 ) = ( 𝐴 − ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) ) )
2	1	oveq2d	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 − ( 𝐴 mod 𝐵 ) ) = ( 𝐴 − ( 𝐴 − ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) ) ) )
3		simpl	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → 𝐴 ∈ ℝ )
4	3	recnd	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → 𝐴 ∈ ℂ )
5		rpcn	⊢ ( 𝐵 ∈ ℝ+ → 𝐵 ∈ ℂ )
6	5	adantl	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → 𝐵 ∈ ℂ )
7		rerpdivcl	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 / 𝐵 ) ∈ ℝ )
8	7	flcld	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ∈ ℤ )
9	8	zcnd	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ∈ ℂ )
10	6 9	mulcld	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) ∈ ℂ )
11	4 10	nncand	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 − ( 𝐴 − ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) ) ) = ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) )
12	2 11	eqtrd	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 − ( 𝐴 mod 𝐵 ) ) = ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) )
13	12	oveq1d	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 − ( 𝐴 mod 𝐵 ) ) / 𝐵 ) = ( ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) / 𝐵 ) )
14		rpne0	⊢ ( 𝐵 ∈ ℝ+ → 𝐵 ≠ 0 )
15	14	adantl	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → 𝐵 ≠ 0 )
16	9 6 15	divcan3d	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐵 · ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) ) / 𝐵 ) = ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) )
17	13 16	eqtrd	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 − ( 𝐴 mod 𝐵 ) ) / 𝐵 ) = ( ⌊ ‘ ( 𝐴 / 𝐵 ) ) )