Metamath Proof Explorer


Theorem lesub1

Description: Subtraction from both sides of 'less than or equal to'. (Contributed by NM, 13-May-2004) (Proof shortened by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion lesub1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴𝐵 ↔ ( 𝐴𝐶 ) ≤ ( 𝐵𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 simp1 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐴 ∈ ℝ )
2 simp3 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℝ )
3 simp2 ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐵 ∈ ℝ )
4 3 2 resubcld ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐵𝐶 ) ∈ ℝ )
5 lesubadd ( ( 𝐴 ∈ ℝ ∧ 𝐶 ∈ ℝ ∧ ( 𝐵𝐶 ) ∈ ℝ ) → ( ( 𝐴𝐶 ) ≤ ( 𝐵𝐶 ) ↔ 𝐴 ≤ ( ( 𝐵𝐶 ) + 𝐶 ) ) )
6 1 2 4 5 syl3anc ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐴𝐶 ) ≤ ( 𝐵𝐶 ) ↔ 𝐴 ≤ ( ( 𝐵𝐶 ) + 𝐶 ) ) )
7 3 recnd ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐵 ∈ ℂ )
8 2 recnd ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → 𝐶 ∈ ℂ )
9 7 8 npcand ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( 𝐵𝐶 ) + 𝐶 ) = 𝐵 )
10 9 breq2d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴 ≤ ( ( 𝐵𝐶 ) + 𝐶 ) ↔ 𝐴𝐵 ) )
11 6 10 bitr2d ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( 𝐴𝐵 ↔ ( 𝐴𝐶 ) ≤ ( 𝐵𝐶 ) ) )