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Metamath Proof Explorer

Theorem le2halvesd

Description: A sum is less than the whole if each term is less than half. (Contributed by Thierry Arnoux, 29-Nov-2017)

Ref Expression
Hypotheses le2halvesd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
le2halvesd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
le2halvesd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℝ )
le2halvesd.4 ⊢ ( 𝜑 → 𝐴 ≤ ( 𝐶 / 2 ) )
le2halvesd.5 ⊢ ( 𝜑 → 𝐵 ≤ ( 𝐶 / 2 ) )
Assertion le2halvesd ( 𝜑 → ( 𝐴 + 𝐵 ) ≤ 𝐶 )

Proof

Step Hyp Ref Expression
1 le2halvesd.1 ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
2 le2halvesd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
3 le2halvesd.3 ⊢ ( 𝜑 → 𝐶 ∈ ℝ )
4 le2halvesd.4 ⊢ ( 𝜑 → 𝐴 ≤ ( 𝐶 / 2 ) )
5 le2halvesd.5 ⊢ ( 𝜑 → 𝐵 ≤ ( 𝐶 / 2 ) )
6 3 rehalfcld ⊢ ( 𝜑 → ( 𝐶 / 2 ) ∈ ℝ )
7 1 2 6 6 4 5 le2addd ⊢ ( 𝜑 → ( 𝐴 + 𝐵 ) ≤ ( ( 𝐶 / 2 ) + ( 𝐶 / 2 ) ) )
8 3 recnd ⊢ ( 𝜑 → 𝐶 ∈ ℂ )
9 8 2halvesd ⊢ ( 𝜑 → ( ( 𝐶 / 2 ) + ( 𝐶 / 2 ) ) = 𝐶 )
10 7 9 breqtrd ⊢ ( 𝜑 → ( 𝐴 + 𝐵 ) ≤ 𝐶 )