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Theorem List for Intuitionistic Logic Explorer - 8101-8200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theorempeano2uz2 8101* Second Peano postulate for upper integers. (Contributed by NM, 3-Oct-2004.)
((A B {x ℤ ∣ Ax}) → (B + 1) {x ℤ ∣ Ax})

Theorempeano5uzti 8102* Peano's inductive postulate for upper integers. (Contributed by NM, 6-Jul-2005.) (Revised by Mario Carneiro, 25-Jul-2013.)
(𝑁 ℤ → ((𝑁 A x A (x + 1) A) → {𝑘 ℤ ∣ 𝑁𝑘} ⊆ A))

Theorempeano5uzi 8103* Peano's inductive postulate for upper integers. (Contributed by NM, 6-Jul-2005.) (Revised by Mario Carneiro, 3-May-2014.)
𝑁        ((𝑁 A x A (x + 1) A) → {𝑘 ℤ ∣ 𝑁𝑘} ⊆ A)

Theoremdfuzi 8104* An expression for the upper integers that start at 𝑁 that is analogous to dfnn2 7677 for positive integers. (Contributed by NM, 6-Jul-2005.) (Proof shortened by Mario Carneiro, 3-May-2014.)
𝑁        {z ℤ ∣ 𝑁z} = {x ∣ (𝑁 x y x (y + 1) x)}

Theoremuzind 8105* Induction on the upper integers that start at 𝑀. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 5-Jul-2005.)
(𝑗 = 𝑀 → (φψ))    &   (𝑗 = 𝑘 → (φχ))    &   (𝑗 = (𝑘 + 1) → (φθ))    &   (𝑗 = 𝑁 → (φτ))    &   (𝑀 ℤ → ψ)    &   ((𝑀 𝑘 𝑀𝑘) → (χθ))       ((𝑀 𝑁 𝑀𝑁) → τ)

Theoremuzind2 8106* Induction on the upper integers that start after an integer 𝑀. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 25-Jul-2005.)
(𝑗 = (𝑀 + 1) → (φψ))    &   (𝑗 = 𝑘 → (φχ))    &   (𝑗 = (𝑘 + 1) → (φθ))    &   (𝑗 = 𝑁 → (φτ))    &   (𝑀 ℤ → ψ)    &   ((𝑀 𝑘 𝑀 < 𝑘) → (χθ))       ((𝑀 𝑁 𝑀 < 𝑁) → τ)

Theoremuzind3 8107* Induction on the upper integers that start at an integer 𝑀. The first four hypotheses give us the substitution instances we need, and the last two are the basis and the induction step. (Contributed by NM, 26-Jul-2005.)
(𝑗 = 𝑀 → (φψ))    &   (𝑗 = 𝑚 → (φχ))    &   (𝑗 = (𝑚 + 1) → (φθ))    &   (𝑗 = 𝑁 → (φτ))    &   (𝑀 ℤ → ψ)    &   ((𝑀 𝑚 {𝑘 ℤ ∣ 𝑀𝑘}) → (χθ))       ((𝑀 𝑁 {𝑘 ℤ ∣ 𝑀𝑘}) → τ)

Theoremnn0ind 8108* Principle of Mathematical Induction (inference schema) on nonnegative integers. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 13-May-2004.)
(x = 0 → (φψ))    &   (x = y → (φχ))    &   (x = (y + 1) → (φθ))    &   (x = A → (φτ))    &   ψ    &   (y 0 → (χθ))       (A 0τ)

Theoremfzind 8109* Induction on the integers from 𝑀 to 𝑁 inclusive . The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by Paul Chapman, 31-Mar-2011.)
(x = 𝑀 → (φψ))    &   (x = y → (φχ))    &   (x = (y + 1) → (φθ))    &   (x = 𝐾 → (φτ))    &   ((𝑀 𝑁 𝑀𝑁) → ψ)    &   (((𝑀 𝑁 ℤ) (y 𝑀y y < 𝑁)) → (χθ))       (((𝑀 𝑁 ℤ) (𝐾 𝑀𝐾 𝐾𝑁)) → τ)

Theoremfnn0ind 8110* Induction on the integers from 0 to 𝑁 inclusive . The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by Paul Chapman, 31-Mar-2011.)
(x = 0 → (φψ))    &   (x = y → (φχ))    &   (x = (y + 1) → (φθ))    &   (x = 𝐾 → (φτ))    &   (𝑁 0ψ)    &   ((𝑁 0 y 0 y < 𝑁) → (χθ))       ((𝑁 0 𝐾 0 𝐾𝑁) → τ)

Theoremnn0ind-raph 8111* Principle of Mathematical Induction (inference schema) on nonnegative integers. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. Raph Levien remarks: "This seems a bit painful. I wonder if an explicit substitution version would be easier." (Contributed by Raph Levien, 10-Apr-2004.)
(x = 0 → (φψ))    &   (x = y → (φχ))    &   (x = (y + 1) → (φθ))    &   (x = A → (φτ))    &   ψ    &   (y 0 → (χθ))       (A 0τ)

Theoremzindd 8112* Principle of Mathematical Induction on all integers, deduction version. The first five hypotheses give the substitutions; the last three are the basis, the induction, and the extension to negative numbers. (Contributed by Paul Chapman, 17-Apr-2009.) (Proof shortened by Mario Carneiro, 4-Jan-2017.)
(x = 0 → (φψ))    &   (x = y → (φχ))    &   (x = (y + 1) → (φτ))    &   (x = -y → (φθ))    &   (x = A → (φη))    &   (ζψ)    &   (ζ → (y 0 → (χτ)))    &   (ζ → (y ℕ → (χθ)))       (ζ → (A ℤ → η))

Theorembtwnz 8113* Any real number can be sandwiched between two integers. Exercise 2 of [Apostol] p. 28. (Contributed by NM, 10-Nov-2004.)
(A ℝ → (x x < A y A < y))

Theoremnn0zd 8114 A positive integer is an integer. (Contributed by Mario Carneiro, 28-May-2016.)
(φA 0)       (φA ℤ)

Theoremnnzd 8115 A nonnegative integer is an integer. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℕ)       (φA ℤ)

Theoremzred 8116 An integer is a real number. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)       (φA ℝ)

Theoremzcnd 8117 An integer is a complex number. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)       (φA ℂ)

Theoremznegcld 8118 Closure law for negative integers. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)       (φ → -A ℤ)

Theorempeano2zd 8119 Deduction from second Peano postulate generalized to integers. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)       (φ → (A + 1) ℤ)

Theoremzaddcld 8120 Closure of addition of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)    &   (φB ℤ)       (φ → (A + B) ℤ)

Theoremzsubcld 8121 Closure of subtraction of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)    &   (φB ℤ)       (φ → (AB) ℤ)

Theoremzmulcld 8122 Closure of multiplication of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(φA ℤ)    &   (φB ℤ)       (φ → (A · B) ℤ)

Theoremzadd2cl 8123 Increasing an integer by 2 results in an integer. (Contributed by Alexander van der Vekens, 16-Sep-2018.)
(𝑁 ℤ → (𝑁 + 2) ℤ)

3.4.9  Decimal arithmetic

Syntaxcdc 8124 Constant used for decimal constructor.
class AB

Definitiondf-dec 8125 Define the "decimal constructor", which is used to build up "decimal integers" or "numeric terms" in base 10. For example, (1000 + 2000) = 3000. (Contributed by Mario Carneiro, 17-Apr-2015.)
AB = ((10 · A) + B)

Theoremdeceq1 8126 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
(A = BA𝐶 = B𝐶)

Theoremdeceq2 8127 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
(A = B𝐶A = 𝐶B)

Theoremdeceq1i 8128 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
A = B       A𝐶 = B𝐶

Theoremdeceq2i 8129 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
A = B       𝐶A = 𝐶B

Theoremdeceq12i 8130 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
A = B    &   𝐶 = 𝐷       A𝐶 = B𝐷

Theoremnumnncl 8131 Closure for a numeral (with units place). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B        ((𝑇 · A) + B)

Theoremnum0u 8132 Add a zero in the units place. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0       (𝑇 · A) = ((𝑇 · A) + 0)

Theoremnum0h 8133 Add a zero in the higher places. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0       A = ((𝑇 · 0) + A)

Theoremnumcl 8134 Closure for a decimal integer (with units place). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0       ((𝑇 · A) + B) 0

Theoremnumsuc 8135 The successor of a decimal integer (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   (B + 1) = 𝐶    &   𝑁 = ((𝑇 · A) + B)       (𝑁 + 1) = ((𝑇 · A) + 𝐶)

Theoremdecnncl 8136 Closure for a numeral. (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0    &   B        AB

Theoremdeccl 8137 Closure for a numeral. (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0    &   B 0       AB 0

Theoremdec0u 8138 Add a zero in the units place. (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0       (10 · A) = A0

Theoremdec0h 8139 Add a zero in the higher places. (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0       A = 0A

Theoremnumnncl2 8140 Closure for a decimal integer (zero units place). (Contributed by Mario Carneiro, 9-Mar-2015.)
𝑇     &   A        ((𝑇 · A) + 0)

Theoremdecnncl2 8141 Closure for a decimal integer (zero units place). (Contributed by Mario Carneiro, 17-Apr-2015.)
A        A0

Theoremnumlt 8142 Comparing two decimal integers (equal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇     &   A 0    &   B 0    &   𝐶     &   B < 𝐶       ((𝑇 · A) + B) < ((𝑇 · A) + 𝐶)

Theoremnumltc 8143 Comparing two decimal integers (unequal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇     &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝐶 < 𝑇    &   A < B       ((𝑇 · A) + 𝐶) < ((𝑇 · B) + 𝐷)

Theoremdeclt 8144 Comparing two decimal integers (equal higher places). (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0    &   B 0    &   𝐶     &   B < 𝐶       AB < A𝐶

Theoremdecltc 8145 Comparing two decimal integers (unequal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝐶 < 10    &   A < B       A𝐶 < B𝐷

Theoremdecsuc 8146 The successor of a decimal integer (no carry). (Contributed by Mario Carneiro, 17-Apr-2015.)
A 0    &   B 0    &   (B + 1) = 𝐶    &   𝑁 = AB       (𝑁 + 1) = A𝐶

Theoremnumlti 8147 Comparing a digit to a decimal integer. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇     &   A     &   B 0    &   𝐶 0    &   𝐶 < 𝑇       𝐶 < ((𝑇 · A) + B)

Theoremdeclti 8148 Comparing a digit to a decimal integer. (Contributed by Mario Carneiro, 18-Feb-2014.)
A     &   B 0    &   𝐶 0    &   𝐶 < 10       𝐶 < AB

Theoremnumsucc 8149 The successor of a decimal integer (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑌 0    &   𝑇 = (𝑌 + 1)    &   A 0    &   (A + 1) = B    &   𝑁 = ((𝑇 · A) + 𝑌)       (𝑁 + 1) = ((𝑇 · B) + 0)

Theoremdecsucc 8150 The successor of a decimal integer (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   (A + 1) = B    &   𝑁 = A9       (𝑁 + 1) = B0

Theorem1e0p1 8151 The successor of zero. (Contributed by Mario Carneiro, 18-Feb-2014.)
1 = (0 + 1)

Theoremdec10p 8152 Ten plus an integer. (Contributed by Mario Carneiro, 19-Apr-2015.)
(10 + A) = 1A

Theoremdec10 8153 The decimal form of 10. NB: In our presentations of large numbers later on, we will use our symbol for 10 at the highest digits when advantageous, because we can use this theorem to convert back to "long form" (where each digit is in the range 0-9) with no extra effort. However, we cannot do this for lower digits while maintaining the ease of use of the decimal system, since it requires nontrivial number knowledge (more than just equality theorems) to convert back. (Contributed by Mario Carneiro, 18-Feb-2014.)
10 = 10

Theoremnumma 8154 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = ((𝑇 · A) + B)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 0    &   ((A · 𝑃) + 𝐶) = 𝐸    &   ((B · 𝑃) + 𝐷) = 𝐹       ((𝑀 · 𝑃) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)

Theoremnummac 8155 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = ((𝑇 · A) + B)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 0    &   𝐹 0    &   𝐺 0    &   ((A · 𝑃) + (𝐶 + 𝐺)) = 𝐸    &   ((B · 𝑃) + 𝐷) = ((𝑇 · 𝐺) + 𝐹)       ((𝑀 · 𝑃) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)

Theoremnumma2c 8156 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = ((𝑇 · A) + B)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 0    &   𝐹 0    &   𝐺 0    &   ((𝑃 · A) + (𝐶 + 𝐺)) = 𝐸    &   ((𝑃 · B) + 𝐷) = ((𝑇 · 𝐺) + 𝐹)       ((𝑃 · 𝑀) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)

Theoremnumadd 8157 Add two decimal integers 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = ((𝑇 · A) + B)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   (A + 𝐶) = 𝐸    &   (B + 𝐷) = 𝐹       (𝑀 + 𝑁) = ((𝑇 · 𝐸) + 𝐹)

Theoremnumaddc 8158 Add two decimal integers 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = ((𝑇 · A) + B)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝐹 0    &   ((A + 𝐶) + 1) = 𝐸    &   (B + 𝐷) = ((𝑇 · 1) + 𝐹)       (𝑀 + 𝑁) = ((𝑇 · 𝐸) + 𝐹)

Theoremnummul1c 8159 The product of a decimal integer with a number. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   𝑃 0    &   A 0    &   B 0    &   𝑁 = ((𝑇 · A) + B)    &   𝐷 0    &   𝐸 0    &   ((A · 𝑃) + 𝐸) = 𝐶    &   (B · 𝑃) = ((𝑇 · 𝐸) + 𝐷)       (𝑁 · 𝑃) = ((𝑇 · 𝐶) + 𝐷)

Theoremnummul2c 8160 The product of a decimal integer with a number (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 0    &   𝑃 0    &   A 0    &   B 0    &   𝑁 = ((𝑇 · A) + B)    &   𝐷 0    &   𝐸 0    &   ((𝑃 · A) + 𝐸) = 𝐶    &   (𝑃 · B) = ((𝑇 · 𝐸) + 𝐷)       (𝑃 · 𝑁) = ((𝑇 · 𝐶) + 𝐷)

Theoremdecma 8161 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   𝑃 0    &   ((A · 𝑃) + 𝐶) = 𝐸    &   ((B · 𝑃) + 𝐷) = 𝐹       ((𝑀 · 𝑃) + 𝑁) = 𝐸𝐹

Theoremdecmac 8162 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   𝑃 0    &   𝐹 0    &   𝐺 0    &   ((A · 𝑃) + (𝐶 + 𝐺)) = 𝐸    &   ((B · 𝑃) + 𝐷) = 𝐺𝐹       ((𝑀 · 𝑃) + 𝑁) = 𝐸𝐹

Theoremdecma2c 8163 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   𝑃 0    &   𝐹 0    &   𝐺 0    &   ((𝑃 · A) + (𝐶 + 𝐺)) = 𝐸    &   ((𝑃 · B) + 𝐷) = 𝐺𝐹       ((𝑃 · 𝑀) + 𝑁) = 𝐸𝐹

Theoremdecadd 8164 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   (A + 𝐶) = 𝐸    &   (B + 𝐷) = 𝐹       (𝑀 + 𝑁) = 𝐸𝐹

Theoremdecaddc 8165 Add two numerals 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   ((A + 𝐶) + 1) = 𝐸    &   𝐹 0    &   (B + 𝐷) = 1𝐹       (𝑀 + 𝑁) = 𝐸𝐹

Theoremdecaddc2 8166 Add two numerals 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝐶 0    &   𝐷 0    &   𝑀 = AB    &   𝑁 = 𝐶𝐷    &   ((A + 𝐶) + 1) = 𝐸    &   (B + 𝐷) = 10       (𝑀 + 𝑁) = 𝐸0

Theoremdecaddi 8167 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝑁 0    &   𝑀 = AB    &   (B + 𝑁) = 𝐶       (𝑀 + 𝑁) = A𝐶

Theoremdecaddci 8168 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝑁 0    &   𝑀 = AB    &   (A + 1) = 𝐷    &   𝐶 0    &   (B + 𝑁) = 1𝐶       (𝑀 + 𝑁) = 𝐷𝐶

Theoremdecaddci2 8169 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
A 0    &   B 0    &   𝑁 0    &   𝑀 = AB    &   (A + 1) = 𝐷    &   (B + 𝑁) = 10       (𝑀 + 𝑁) = 𝐷0

Theoremdecmul1c 8170 The product of a numeral with a number. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑃 0    &   A 0    &   B 0    &   𝑁 = AB    &   𝐷 0    &   𝐸 0    &   ((A · 𝑃) + 𝐸) = 𝐶    &   (B · 𝑃) = 𝐸𝐷       (𝑁 · 𝑃) = 𝐶𝐷

Theoremdecmul2c 8171 The product of a numeral with a number (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑃 0    &   A 0    &   B 0    &   𝑁 = AB    &   𝐷 0    &   𝐸 0    &   ((𝑃 · A) + 𝐸) = 𝐶    &   (𝑃 · B) = 𝐸𝐷       (𝑃 · 𝑁) = 𝐶𝐷

Theorem6p5lem 8172 Lemma for 6p5e11 8173 and related theorems. (Contributed by Mario Carneiro, 19-Apr-2015.)
A 0    &   𝐷 0    &   𝐸 0    &   B = (𝐷 + 1)    &   𝐶 = (𝐸 + 1)    &   (A + 𝐷) = 1𝐸       (A + B) = 1𝐶

Theorem6p5e11 8173 6 + 5 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(6 + 5) = 11

Theorem6p6e12 8174 6 + 6 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(6 + 6) = 12

Theorem7p4e11 8175 7 + 4 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 4) = 11

Theorem7p5e12 8176 7 + 5 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 5) = 12

Theorem7p6e13 8177 7 + 6 = 13. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 6) = 13

Theorem7p7e14 8178 7 + 7 = 14. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 7) = 14

Theorem8p3e11 8179 8 + 3 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 3) = 11

Theorem8p4e12 8180 8 + 4 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 4) = 12

Theorem8p5e13 8181 8 + 5 = 13. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 5) = 13

Theorem8p6e14 8182 8 + 6 = 14. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 6) = 14

Theorem8p7e15 8183 8 + 7 = 15. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 7) = 15

Theorem8p8e16 8184 8 + 8 = 16. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 8) = 16

Theorem9p2e11 8185 9 + 2 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 2) = 11

Theorem9p3e12 8186 9 + 3 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 3) = 12

Theorem9p4e13 8187 9 + 4 = 13. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 4) = 13

Theorem9p5e14 8188 9 + 5 = 14. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 5) = 14

Theorem9p6e15 8189 9 + 6 = 15. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 6) = 15

Theorem9p7e16 8190 9 + 7 = 16. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 7) = 16

Theorem9p8e17 8191 9 + 8 = 17. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 8) = 17

Theorem9p9e18 8192 9 + 9 = 18. (Contributed by Mario Carneiro, 19-Apr-2015.)
(9 + 9) = 18

Theorem10p10e20 8193 10 + 10 = 20. (Contributed by Mario Carneiro, 19-Apr-2015.)
(10 + 10) = 20

Theorem4t3lem 8194 Lemma for 4t3e12 8195 and related theorems. (Contributed by Mario Carneiro, 19-Apr-2015.)
A 0    &   B 0    &   𝐶 = (B + 1)    &   (A · B) = 𝐷    &   (𝐷 + A) = 𝐸       (A · 𝐶) = 𝐸

Theorem4t3e12 8195 4 times 3 equals 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(4 · 3) = 12

Theorem4t4e16 8196 4 times 4 equals 16. (Contributed by Mario Carneiro, 19-Apr-2015.)
(4 · 4) = 16

Theorem5t3e15 8197 5 times 3 equals 15. (Contributed by Mario Carneiro, 19-Apr-2015.)
(5 · 3) = 15

Theorem5t4e20 8198 5 times 4 equals 20. (Contributed by Mario Carneiro, 19-Apr-2015.)
(5 · 4) = 20

Theorem5t5e25 8199 5 times 5 equals 25. (Contributed by Mario Carneiro, 19-Apr-2015.)
(5 · 5) = 25

Theorem6t2e12 8200 6 times 2 equals 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(6 · 2) = 12

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