Theorem modid2

Description: Identity law for modulo. (Contributed by NM, 29-Dec-2008)

	Ref	Expression
Assertion	modid2	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 mod 𝐵 ) = 𝐴 ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) ) )

Proof

Step	Hyp	Ref	Expression
1		modge0	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → 0 ≤ ( 𝐴 mod 𝐵 ) )
2		modlt	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 𝐴 mod 𝐵 ) < 𝐵 )
3	1 2	jca	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( 0 ≤ ( 𝐴 mod 𝐵 ) ∧ ( 𝐴 mod 𝐵 ) < 𝐵 ) )
4		breq2	⊢ ( ( 𝐴 mod 𝐵 ) = 𝐴 → ( 0 ≤ ( 𝐴 mod 𝐵 ) ↔ 0 ≤ 𝐴 ) )
5		breq1	⊢ ( ( 𝐴 mod 𝐵 ) = 𝐴 → ( ( 𝐴 mod 𝐵 ) < 𝐵 ↔ 𝐴 < 𝐵 ) )
6	4 5	anbi12d	⊢ ( ( 𝐴 mod 𝐵 ) = 𝐴 → ( ( 0 ≤ ( 𝐴 mod 𝐵 ) ∧ ( 𝐴 mod 𝐵 ) < 𝐵 ) ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) ) )
7	3 6	syl5ibcom	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 mod 𝐵 ) = 𝐴 → ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) ) )
8		modid	⊢ ( ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) ∧ ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) ) → ( 𝐴 mod 𝐵 ) = 𝐴 )
9	8	ex	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) → ( 𝐴 mod 𝐵 ) = 𝐴 ) )
10	7 9	impbid	⊢ ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ+ ) → ( ( 𝐴 mod 𝐵 ) = 𝐴 ↔ ( 0 ≤ 𝐴 ∧ 𝐴 < 𝐵 ) ) )