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Theorem List for Metamath Proof Explorer - 10501-10600   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremsubdi 10501 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by NM, 18-Nov-2004.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴 · (𝐵𝐶)) = ((𝐴 · 𝐵) − (𝐴 · 𝐶)))

Theoremsubdir 10502 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by NM, 30-Dec-2005.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → ((𝐴𝐵) · 𝐶) = ((𝐴 · 𝐶) − (𝐵 · 𝐶)))

Theoremine0 10503 The imaginary unit i is not zero. (Contributed by NM, 6-May-1999.)
i ≠ 0

Theoremmulneg1 10504 Product with negative is negative of product. Theorem I.12 of [Apostol] p. 18. (Contributed by NM, 14-May-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (-𝐴 · 𝐵) = -(𝐴 · 𝐵))

Theoremmulneg2 10505 The product with a negative is the negative of the product. (Contributed by NM, 30-Jul-2004.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 · -𝐵) = -(𝐴 · 𝐵))

Theoremmulneg12 10506 Swap the negative sign in a product. (Contributed by NM, 30-Jul-2004.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (-𝐴 · 𝐵) = (𝐴 · -𝐵))

Theoremmul2neg 10507 Product of two negatives. Theorem I.12 of [Apostol] p. 18. (Contributed by NM, 30-Jul-2004.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (-𝐴 · -𝐵) = (𝐴 · 𝐵))

Theoremsubmul2 10508 Convert a subtraction to addition using multiplication by a negative. (Contributed by NM, 2-Feb-2007.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (𝐴 − (𝐵 · 𝐶)) = (𝐴 + (𝐵 · -𝐶)))

Theoremmulm1 10509 Product with minus one is negative. (Contributed by NM, 16-Nov-1999.)
(𝐴 ∈ ℂ → (-1 · 𝐴) = -𝐴)

Theoremaddneg1mul 10510 Addition with product with minus one is a subtraction. (Contributed by AV, 18-Oct-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (𝐴 + (-1 · 𝐵)) = (𝐴𝐵))

Theoremmulsub 10511 Product of two differences. (Contributed by NM, 14-Jan-2006.)
(((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴𝐵) · (𝐶𝐷)) = (((𝐴 · 𝐶) + (𝐷 · 𝐵)) − ((𝐴 · 𝐷) + (𝐶 · 𝐵))))

Theoremmulsub2 10512 Swap the order of subtraction in a multiplication. (Contributed by Scott Fenton, 24-Jun-2013.)
(((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐷 ∈ ℂ)) → ((𝐴𝐵) · (𝐶𝐷)) = ((𝐵𝐴) · (𝐷𝐶)))

Theoremmulm1i 10513 Product with minus one is negative. (Contributed by NM, 31-Jul-1999.)
𝐴 ∈ ℂ       (-1 · 𝐴) = -𝐴

Theoremmulneg1i 10514 Product with negative is negative of product. Theorem I.12 of [Apostol] p. 18. (Contributed by NM, 10-Feb-1995.) (Revised by Mario Carneiro, 27-May-2016.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ       (-𝐴 · 𝐵) = -(𝐴 · 𝐵)

Theoremmulneg2i 10515 Product with negative is negative of product. (Contributed by NM, 31-Jul-1999.) (Revised by Mario Carneiro, 27-May-2016.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ       (𝐴 · -𝐵) = -(𝐴 · 𝐵)

Theoremmul2negi 10516 Product of two negatives. Theorem I.12 of [Apostol] p. 18. (Contributed by NM, 14-Feb-1995.) (Revised by Mario Carneiro, 27-May-2016.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ       (-𝐴 · -𝐵) = (𝐴 · 𝐵)

Theoremsubdii 10517 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by NM, 26-Nov-1994.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ    &   𝐶 ∈ ℂ       (𝐴 · (𝐵𝐶)) = ((𝐴 · 𝐵) − (𝐴 · 𝐶))

Theoremsubdiri 10518 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by NM, 8-May-1999.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ    &   𝐶 ∈ ℂ       ((𝐴𝐵) · 𝐶) = ((𝐴 · 𝐶) − (𝐵 · 𝐶))

Theoremmuladdi 10519 Product of two sums. (Contributed by NM, 17-May-1999.)
𝐴 ∈ ℂ    &   𝐵 ∈ ℂ    &   𝐶 ∈ ℂ    &   𝐷 ∈ ℂ       ((𝐴 + 𝐵) · (𝐶 + 𝐷)) = (((𝐴 · 𝐶) + (𝐷 · 𝐵)) + ((𝐴 · 𝐷) + (𝐶 · 𝐵)))

Theoremmulm1d 10520 Product with minus one is negative. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → (-1 · 𝐴) = -𝐴)

Theoremmulneg1d 10521 Product with negative is negative of product. Theorem I.12 of [Apostol] p. 18. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (-𝐴 · 𝐵) = -(𝐴 · 𝐵))

Theoremmulneg2d 10522 Product with negative is negative of product. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (𝐴 · -𝐵) = -(𝐴 · 𝐵))

Theoremmul2negd 10523 Product of two negatives. Theorem I.12 of [Apostol] p. 18. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (-𝐴 · -𝐵) = (𝐴 · 𝐵))

Theoremsubdid 10524 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)       (𝜑 → (𝐴 · (𝐵𝐶)) = ((𝐴 · 𝐵) − (𝐴 · 𝐶)))

Theoremsubdird 10525 Distribution of multiplication over subtraction. Theorem I.5 of [Apostol] p. 18. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)       (𝜑 → ((𝐴𝐵) · 𝐶) = ((𝐴 · 𝐶) − (𝐵 · 𝐶)))

Theoremsubdir2d 10526 Distribution of multiplication over subtraction. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)       (𝜑 → (𝐶 · (𝐴𝐵)) = ((𝐶 · 𝐴) − (𝐶 · 𝐵)))

Theoremmuladdd 10527 Product of two sums. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → ((𝐴 + 𝐵) · (𝐶 + 𝐷)) = (((𝐴 · 𝐶) + (𝐷 · 𝐵)) + ((𝐴 · 𝐷) + (𝐶 · 𝐵))))

Theoremmulsubd 10528 Product of two differences. (Contributed by Mario Carneiro, 27-May-2016.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → ((𝐴𝐵) · (𝐶𝐷)) = (((𝐴 · 𝐶) + (𝐷 · 𝐵)) − ((𝐴 · 𝐷) + (𝐶 · 𝐵))))

Theoremmuls1d 10529 Multiplication by one minus a number. (Contributed by Scott Fenton, 23-Dec-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (𝐴 · (𝐵 − 1)) = ((𝐴 · 𝐵) − 𝐴))

Theoremmulsubfacd 10530 Multiplication followed by the subtraction of a factor. (Contributed by Alexander van der Vekens, 28-Aug-2018.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → ((𝐴 · 𝐵) − 𝐵) = ((𝐴 − 1) · 𝐵))

5.3.4  Ordering on reals (cont.)

Theoremgt0ne0 10531 Positive implies nonzero. (Contributed by NM, 3-Oct-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 0 < 𝐴) → 𝐴 ≠ 0)

Theoremlt0ne0 10532 A number which is less than zero is not zero. (Contributed by Stefan O'Rear, 13-Sep-2014.)
((𝐴 ∈ ℝ ∧ 𝐴 < 0) → 𝐴 ≠ 0)

Theoremltadd1 10533 Addition to both sides of 'less than'. (Contributed by NM, 12-Nov-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴 < 𝐵 ↔ (𝐴 + 𝐶) < (𝐵 + 𝐶)))

Theoremleadd1 10534 Addition to both sides of 'less than or equal to'. (Contributed by NM, 18-Oct-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴𝐵 ↔ (𝐴 + 𝐶) ≤ (𝐵 + 𝐶)))

Theoremleadd2 10535 Addition to both sides of 'less than or equal to'. (Contributed by NM, 26-Oct-1999.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴𝐵 ↔ (𝐶 + 𝐴) ≤ (𝐶 + 𝐵)))

Theoremltsubadd 10536 'Less than' relationship between subtraction and addition. (Contributed by NM, 21-Jan-1997.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) < 𝐶𝐴 < (𝐶 + 𝐵)))

Theoremltsubadd2 10537 'Less than' relationship between subtraction and addition. (Contributed by NM, 21-Jan-1997.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) < 𝐶𝐴 < (𝐵 + 𝐶)))

Theoremlesubadd 10538 'Less than or equal to' relationship between subtraction and addition. (Contributed by NM, 17-Nov-2004.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) ≤ 𝐶𝐴 ≤ (𝐶 + 𝐵)))

Theoremlesubadd2 10539 'Less than or equal to' relationship between subtraction and addition. (Contributed by NM, 10-Aug-1999.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) ≤ 𝐶𝐴 ≤ (𝐵 + 𝐶)))

Theoremltaddsub 10540 'Less than' relationship between addition and subtraction. (Contributed by NM, 17-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴 + 𝐵) < 𝐶𝐴 < (𝐶𝐵)))

Theoremltaddsub2 10541 'Less than' relationship between addition and subtraction. (Contributed by NM, 17-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴 + 𝐵) < 𝐶𝐵 < (𝐶𝐴)))

Theoremleaddsub 10542 'Less than or equal to' relationship between addition and subtraction. (Contributed by NM, 6-Apr-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴 + 𝐵) ≤ 𝐶𝐴 ≤ (𝐶𝐵)))

Theoremleaddsub2 10543 'Less than or equal to' relationship between and addition and subtraction. (Contributed by NM, 6-Apr-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴 + 𝐵) ≤ 𝐶𝐵 ≤ (𝐶𝐴)))

Theoremsuble 10544 Swap subtrahends in an inequality. (Contributed by NM, 29-Sep-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) ≤ 𝐶 ↔ (𝐴𝐶) ≤ 𝐵))

Theoremlesub 10545 Swap subtrahends in an inequality. (Contributed by NM, 29-Sep-2005.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴 ≤ (𝐵𝐶) ↔ 𝐶 ≤ (𝐵𝐴)))

Theoremltsub23 10546 'Less than' relationship between subtraction and addition. (Contributed by NM, 4-Oct-1999.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((𝐴𝐵) < 𝐶 ↔ (𝐴𝐶) < 𝐵))

Theoremltsub13 10547 'Less than' relationship between subtraction and addition. (Contributed by NM, 17-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴 < (𝐵𝐶) ↔ 𝐶 < (𝐵𝐴)))

Theoremle2add 10548 Adding both sides of two 'less than or equal to' relations. (Contributed by NM, 17-Apr-2005.) (Proof shortened by Mario Carneiro, 27-May-2016.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴𝐶𝐵𝐷) → (𝐴 + 𝐵) ≤ (𝐶 + 𝐷)))

Theoremltleadd 10549 Adding both sides of two orderings. (Contributed by NM, 23-Dec-2007.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴 < 𝐶𝐵𝐷) → (𝐴 + 𝐵) < (𝐶 + 𝐷)))

Theoremleltadd 10550 Adding both sides of two orderings. (Contributed by NM, 15-Aug-2008.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴𝐶𝐵 < 𝐷) → (𝐴 + 𝐵) < (𝐶 + 𝐷)))

Theoremlt2add 10551 Adding both sides of two 'less than' relations. Theorem I.25 of [Apostol] p. 20. (Contributed by NM, 15-Aug-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴 < 𝐶𝐵 < 𝐷) → (𝐴 + 𝐵) < (𝐶 + 𝐷)))

Theoremaddgt0 10552 The sum of 2 positive numbers is positive. (Contributed by NM, 1-Jun-2005.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (0 < 𝐴 ∧ 0 < 𝐵)) → 0 < (𝐴 + 𝐵))

Theoremaddgegt0 10553 The sum of nonnegative and positive numbers is positive. (Contributed by NM, 28-Dec-2005.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (0 ≤ 𝐴 ∧ 0 < 𝐵)) → 0 < (𝐴 + 𝐵))

Theoremaddgtge0 10554 The sum of nonnegative and positive numbers is positive. (Contributed by NM, 28-Dec-2005.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (0 < 𝐴 ∧ 0 ≤ 𝐵)) → 0 < (𝐴 + 𝐵))

Theoremaddge0 10555 The sum of 2 nonnegative numbers is nonnegative. (Contributed by NM, 17-Mar-2005.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (0 ≤ 𝐴 ∧ 0 ≤ 𝐵)) → 0 ≤ (𝐴 + 𝐵))

Theoremltaddpos 10556 Adding a positive number to another number increases it. (Contributed by NM, 17-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 < 𝐴𝐵 < (𝐵 + 𝐴)))

Theoremltaddpos2 10557 Adding a positive number to another number increases it. (Contributed by NM, 8-Apr-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 < 𝐴𝐵 < (𝐴 + 𝐵)))

Theoremltsubpos 10558 Subtracting a positive number from another number decreases it. (Contributed by NM, 17-Nov-2004.) (Proof shortened by Andrew Salmon, 19-Nov-2011.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 < 𝐴 ↔ (𝐵𝐴) < 𝐵))

Theoremposdif 10559 Comparison of two numbers whose difference is positive. (Contributed by NM, 17-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < 𝐵 ↔ 0 < (𝐵𝐴)))

Theoremlesub1 10560 Subtraction from both sides of 'less than or equal to'. (Contributed by NM, 13-May-2004.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴𝐵 ↔ (𝐴𝐶) ≤ (𝐵𝐶)))

Theoremlesub2 10561 Subtraction of both sides of 'less than or equal to'. (Contributed by NM, 29-Sep-2005.) (Revised by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴𝐵 ↔ (𝐶𝐵) ≤ (𝐶𝐴)))

Theoremltsub1 10562 Subtraction from both sides of 'less than'. (Contributed by FL, 3-Jan-2008.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴 < 𝐵 ↔ (𝐴𝐶) < (𝐵𝐶)))

Theoremltsub2 10563 Subtraction of both sides of 'less than'. (Contributed by NM, 29-Sep-2005.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐴 < 𝐵 ↔ (𝐶𝐵) < (𝐶𝐴)))

Theoremlt2sub 10564 Subtracting both sides of two 'less than' relations. (Contributed by Mario Carneiro, 14-Apr-2016.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴 < 𝐶𝐷 < 𝐵) → (𝐴𝐵) < (𝐶𝐷)))

Theoremle2sub 10565 Subtracting both sides of two 'less than or equal to' relations. (Contributed by Mario Carneiro, 14-Apr-2016.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) ∧ (𝐶 ∈ ℝ ∧ 𝐷 ∈ ℝ)) → ((𝐴𝐶𝐷𝐵) → (𝐴𝐵) ≤ (𝐶𝐷)))

Theoremltneg 10566 Negative of both sides of 'less than'. Theorem I.23 of [Apostol] p. 20. (Contributed by NM, 27-Aug-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < 𝐵 ↔ -𝐵 < -𝐴))

Theoremltnegcon1 10567 Contraposition of negative in 'less than'. (Contributed by NM, 8-Nov-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (-𝐴 < 𝐵 ↔ -𝐵 < 𝐴))

Theoremltnegcon2 10568 Contraposition of negative in 'less than'. (Contributed by Mario Carneiro, 25-Feb-2015.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < -𝐵𝐵 < -𝐴))

Theoremleneg 10569 Negative of both sides of 'less than or equal to'. (Contributed by NM, 12-Sep-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴𝐵 ↔ -𝐵 ≤ -𝐴))

Theoremlenegcon1 10570 Contraposition of negative in 'less than or equal to'. (Contributed by NM, 10-May-2004.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (-𝐴𝐵 ↔ -𝐵𝐴))

Theoremlenegcon2 10571 Contraposition of negative in 'less than or equal to'. (Contributed by NM, 8-Oct-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 ≤ -𝐵𝐵 ≤ -𝐴))

Theoremlt0neg1 10572 Comparison of a number and its negative to zero. Theorem I.23 of [Apostol] p. 20. (Contributed by NM, 14-May-1999.)
(𝐴 ∈ ℝ → (𝐴 < 0 ↔ 0 < -𝐴))

Theoremlt0neg2 10573 Comparison of a number and its negative to zero. (Contributed by NM, 10-May-2004.)
(𝐴 ∈ ℝ → (0 < 𝐴 ↔ -𝐴 < 0))

Theoremle0neg1 10574 Comparison of a number and its negative to zero. (Contributed by NM, 10-May-2004.)
(𝐴 ∈ ℝ → (𝐴 ≤ 0 ↔ 0 ≤ -𝐴))

Theoremle0neg2 10575 Comparison of a number and its negative to zero. (Contributed by NM, 24-Aug-1999.)
(𝐴 ∈ ℝ → (0 ≤ 𝐴 ↔ -𝐴 ≤ 0))

Theoremaddge01 10576 A number is less than or equal to itself plus a nonnegative number. (Contributed by NM, 21-Feb-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 ≤ 𝐵𝐴 ≤ (𝐴 + 𝐵)))

Theoremaddge02 10577 A number is less than or equal to itself plus a nonnegative number. (Contributed by NM, 27-Jul-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 ≤ 𝐵𝐴 ≤ (𝐵 + 𝐴)))

Theoremadd20 10578 Two nonnegative numbers are zero iff their sum is zero. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 27-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((𝐴 + 𝐵) = 0 ↔ (𝐴 = 0 ∧ 𝐵 = 0)))

Theoremsubge0 10579 Nonnegative subtraction. (Contributed by NM, 14-Mar-2005.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 ≤ (𝐴𝐵) ↔ 𝐵𝐴))

Theoremsuble0 10580 Nonpositive subtraction. (Contributed by NM, 20-Mar-2008.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴𝐵) ≤ 0 ↔ 𝐴𝐵))

Theoremleaddle0 10581 The sum of a real number and a second real number is less than the real number iff the second real number is negative. (Contributed by Alexander van der Vekens, 30-May-2018.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴 + 𝐵) ≤ 𝐴𝐵 ≤ 0))

Theoremsubge02 10582 Nonnegative subtraction. (Contributed by NM, 27-Jul-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (0 ≤ 𝐵 ↔ (𝐴𝐵) ≤ 𝐴))

Theoremlesub0 10583 Lemma to show a nonnegative number is zero. (Contributed by NM, 8-Oct-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((0 ≤ 𝐴𝐵 ≤ (𝐵𝐴)) ↔ 𝐴 = 0))

Theoremmulge0 10584 The product of two nonnegative numbers is nonnegative. (Contributed by NM, 8-Oct-1999.) (Revised by Mario Carneiro, 27-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → 0 ≤ (𝐴 · 𝐵))

Theoremmullt0 10585 The product of two negative numbers is positive. (Contributed by Jeff Hankins, 8-Jun-2009.)
(((𝐴 ∈ ℝ ∧ 𝐴 < 0) ∧ (𝐵 ∈ ℝ ∧ 𝐵 < 0)) → 0 < (𝐴 · 𝐵))

Theoremmsqgt0 10586 A nonzero square is positive. Theorem I.20 of [Apostol] p. 20. (Contributed by NM, 6-May-1999.) (Proof shortened by Mario Carneiro, 27-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐴 ≠ 0) → 0 < (𝐴 · 𝐴))

Theoremmsqge0 10587 A square is nonnegative. (Contributed by NM, 23-May-2007.) (Revised by Mario Carneiro, 27-May-2016.)
(𝐴 ∈ ℝ → 0 ≤ (𝐴 · 𝐴))

Theorem0lt1 10588 0 is less than 1. Theorem I.21 of [Apostol] p. 20. (Contributed by NM, 17-Jan-1997.)
0 < 1

Theorem0le1 10589 0 is less than or equal to 1. (Contributed by Mario Carneiro, 29-Apr-2015.)
0 ≤ 1

Theoremrelin01 10590 An interval law for less than or equal. (Contributed by Scott Fenton, 27-Jun-2013.)
(𝐴 ∈ ℝ → (𝐴 ≤ 0 ∨ (0 ≤ 𝐴𝐴 ≤ 1) ∨ 1 ≤ 𝐴))

Theoremltordlem 10591* Lemma for ltord1 10592. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐴 < 𝐵))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶 < 𝐷𝑀 < 𝑁))

Theoremltord1 10592* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐴 < 𝐵))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶 < 𝐷𝑀 < 𝑁))

Theoremleord1 10593* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐴 < 𝐵))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶𝐷𝑀𝑁))

Theoremeqord1 10594* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐴 < 𝐵))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶 = 𝐷𝑀 = 𝑁))

Theoremltord2 10595* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐵 < 𝐴))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶 < 𝐷𝑁 < 𝑀))

Theoremleord2 10596* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐵 < 𝐴))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶𝐷𝑁𝑀))

Theoremeqord2 10597* Infer an ordering relation from a proof in only one direction. (Contributed by Mario Carneiro, 14-Jun-2014.)
(𝑥 = 𝑦𝐴 = 𝐵)    &   (𝑥 = 𝐶𝐴 = 𝑀)    &   (𝑥 = 𝐷𝐴 = 𝑁)    &   𝑆 ⊆ ℝ    &   ((𝜑𝑥𝑆) → 𝐴 ∈ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 < 𝑦𝐵 < 𝐴))       ((𝜑 ∧ (𝐶𝑆𝐷𝑆)) → (𝐶 = 𝐷𝑀 = 𝑁))

Theoremwloglei 10598* Form of wlogle 10599 where both sides of the equivalence are proven rather than showing that they are equivalent to each other. (Contributed by Mario Carneiro, 9-Mar-2015.)
((𝑧 = 𝑥𝑤 = 𝑦) → (𝜓𝜒))    &   ((𝑧 = 𝑦𝑤 = 𝑥) → (𝜓𝜃))    &   (𝜑𝑆 ⊆ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑥𝑦)) → 𝜃)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑥𝑦)) → 𝜒)       ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → 𝜒)

Theoremwlogle 10599* If the predicate 𝜒(𝑥, 𝑦) is symmetric under interchange of 𝑥, 𝑦, then "without loss of generality" we can assume that 𝑥𝑦. (Contributed by Mario Carneiro, 18-Aug-2014.) (Revised by Mario Carneiro, 11-Sep-2014.)
((𝑧 = 𝑥𝑤 = 𝑦) → (𝜓𝜒))    &   ((𝑧 = 𝑦𝑤 = 𝑥) → (𝜓𝜃))    &   (𝜑𝑆 ⊆ ℝ)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝜒𝜃))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑥𝑦)) → 𝜒)       ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → 𝜒)

Theoremleidi 10600 'Less than or equal to' is reflexive. (Contributed by NM, 18-Aug-1999.)
𝐴 ∈ ℝ       𝐴𝐴

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