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Theorem List for Intuitionistic Logic Explorer - 10101-10200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremiseqf1olemqval 10101* Lemma for seq3f1o 10118. Value of the function 𝑄. (Contributed by Jim Kingdon, 28-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))       (𝜑 → (𝑄𝐴) = if(𝐴 ∈ (𝐾...(𝐽𝐾)), if(𝐴 = 𝐾, 𝐾, (𝐽‘(𝐴 − 1))), (𝐽𝐴)))

Theoremiseqf1olemnab 10102* Lemma for seq3f1o 10118. (Contributed by Jim Kingdon, 27-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   (𝜑𝐵 ∈ (𝑀...𝑁))    &   (𝜑 → (𝑄𝐴) = (𝑄𝐵))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))       (𝜑 → ¬ (𝐴 ∈ (𝐾...(𝐽𝐾)) ∧ ¬ 𝐵 ∈ (𝐾...(𝐽𝐾))))

Theoremiseqf1olemab 10103* Lemma for seq3f1o 10118. (Contributed by Jim Kingdon, 27-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   (𝜑𝐵 ∈ (𝑀...𝑁))    &   (𝜑 → (𝑄𝐴) = (𝑄𝐵))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   (𝜑𝐴 ∈ (𝐾...(𝐽𝐾)))    &   (𝜑𝐵 ∈ (𝐾...(𝐽𝐾)))       (𝜑𝐴 = 𝐵)

Theoremiseqf1olemnanb 10104* Lemma for seq3f1o 10118. (Contributed by Jim Kingdon, 27-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   (𝜑𝐵 ∈ (𝑀...𝑁))    &   (𝜑 → (𝑄𝐴) = (𝑄𝐵))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   (𝜑 → ¬ 𝐴 ∈ (𝐾...(𝐽𝐾)))    &   (𝜑 → ¬ 𝐵 ∈ (𝐾...(𝐽𝐾)))       (𝜑𝐴 = 𝐵)

Theoremiseqf1olemqf 10105* Lemma for seq3f1o 10118. Domain and codomain of 𝑄. (Contributed by Jim Kingdon, 26-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))       (𝜑𝑄:(𝑀...𝑁)⟶(𝑀...𝑁))

Theoremiseqf1olemmo 10106* Lemma for seq3f1o 10118. Showing that 𝑄 is one-to-one. (Contributed by Jim Kingdon, 27-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   (𝜑𝐵 ∈ (𝑀...𝑁))    &   (𝜑 → (𝑄𝐴) = (𝑄𝐵))       (𝜑𝐴 = 𝐵)

Theoremiseqf1olemqf1o 10107* Lemma for seq3f1o 10118. 𝑄 is a permutation of (𝑀...𝑁). 𝑄 is formed from the constant portion of 𝐽, followed by the single element 𝐾 (at position 𝐾), followed by the rest of J (with the 𝐾 deleted and the elements before 𝐾 moved one position later to fill the gap). (Contributed by Jim Kingdon, 21-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))       (𝜑𝑄:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))

Theoremiseqf1olemqk 10108* Lemma for seq3f1o 10118. 𝑄 is constant for one more position than 𝐽 is. (Contributed by Jim Kingdon, 21-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   (𝜑 → ∀𝑥 ∈ (𝑀..^𝐾)(𝐽𝑥) = 𝑥)       (𝜑 → ∀𝑥 ∈ (𝑀...𝐾)(𝑄𝑥) = 𝑥)

Theoremiseqf1olemjpcl 10109* Lemma for seq3f1o 10118. A closure lemma involving 𝐽 and 𝑃. (Contributed by Jim Kingdon, 29-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐽 / 𝑓𝑃𝑥) ∈ 𝑆)

Theoremiseqf1olemqpcl 10110* Lemma for seq3f1o 10118. A closure lemma involving 𝑄 and 𝑃. (Contributed by Jim Kingdon, 29-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝑄 / 𝑓𝑃𝑥) ∈ 𝑆)

Theoremiseqf1olemfvp 10111* Lemma for seq3f1o 10118. (Contributed by Jim Kingdon, 30-Aug-2022.)
(𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝑇:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑𝐴 ∈ (𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → (𝑇 / 𝑓𝑃𝐴) = (𝐺‘(𝑇𝐴)))

Theoremseq3f1olemqsumkj 10112* Lemma for seq3f1o 10118. 𝑄 gives the same sum as 𝐽 in the range (𝐾...(𝐽𝐾)). (Contributed by Jim Kingdon, 29-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   (𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑 → ∀𝑥 ∈ (𝑀..^𝐾)(𝐽𝑥) = 𝑥)    &   (𝜑𝐾 ≠ (𝐽𝐾))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → (seq𝐾( + , 𝐽 / 𝑓𝑃)‘(𝐽𝐾)) = (seq𝐾( + , 𝑄 / 𝑓𝑃)‘(𝐽𝐾)))

Theoremseq3f1olemqsumk 10113* Lemma for seq3f1o 10118. 𝑄 gives the same sum as 𝐽 in the range (𝐾...𝑁). (Contributed by Jim Kingdon, 22-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   (𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑 → ∀𝑥 ∈ (𝑀..^𝐾)(𝐽𝑥) = 𝑥)    &   (𝜑𝐾 ≠ (𝐽𝐾))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → (seq𝐾( + , 𝐽 / 𝑓𝑃)‘𝑁) = (seq𝐾( + , 𝑄 / 𝑓𝑃)‘𝑁))

Theoremseq3f1olemqsum 10114* Lemma for seq3f1o 10118. 𝑄 gives the same sum as 𝐽. (Contributed by Jim Kingdon, 21-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   (𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑 → ∀𝑥 ∈ (𝑀..^𝐾)(𝐽𝑥) = 𝑥)    &   (𝜑𝐾 ≠ (𝐽𝐾))    &   𝑄 = (𝑢 ∈ (𝑀...𝑁) ↦ if(𝑢 ∈ (𝐾...(𝐽𝐾)), if(𝑢 = 𝐾, 𝐾, (𝐽‘(𝑢 − 1))), (𝐽𝑢)))    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → (seq𝑀( + , 𝐽 / 𝑓𝑃)‘𝑁) = (seq𝑀( + , 𝑄 / 𝑓𝑃)‘𝑁))

Theoremseq3f1olemstep 10115* Lemma for seq3f1o 10118. Given a permutation which is constant up to a point, supply a new one which is constant for one more position. (Contributed by Jim Kingdon, 19-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   (𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑𝐽:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   (𝜑 → ∀𝑥 ∈ (𝑀..^𝐾)(𝐽𝑥) = 𝑥)    &   (𝜑 → (seq𝑀( + , 𝐽 / 𝑓𝑃)‘𝑁) = (seq𝑀( + , 𝐿)‘𝑁))    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → ∃𝑓(𝑓:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁) ∧ ∀𝑥 ∈ (𝑀...𝐾)(𝑓𝑥) = 𝑥 ∧ (seq𝑀( + , 𝑃)‘𝑁) = (seq𝑀( + , 𝐿)‘𝑁)))

Theoremseq3f1olemp 10116* Lemma for seq3f1o 10118. Existence of a constant permutation of (𝑀...𝑁) which leads to the same sum as the permutation 𝐹 itself. (Contributed by Jim Kingdon, 18-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   𝐿 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝐹𝑥)), (𝐺𝑀)))    &   𝑃 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝑓𝑥)), (𝐺𝑀)))       (𝜑 → ∃𝑓(𝑓:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁) ∧ ∀𝑥 ∈ (𝑀...𝑁)(𝑓𝑥) = 𝑥 ∧ (seq𝑀( + , 𝑃)‘𝑁) = (seq𝑀( + , 𝐿)‘𝑁)))

Theoremseq3f1oleml 10117* Lemma for seq3f1o 10118. This is more or less the result, but stated in terms of 𝐹 and 𝐺 without 𝐻. 𝐿 and 𝐻 may differ in terms of what happens to terms after 𝑁. The terms after 𝑁 don't matter for the value at 𝑁 but we need some definition given the way our theorems concerning seq work. (Contributed by Jim Kingdon, 17-Aug-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   𝐿 = (𝑥 ∈ (ℤ𝑀) ↦ if(𝑥𝑁, (𝐺‘(𝐹𝑥)), (𝐺𝑀)))       (𝜑 → (seq𝑀( + , 𝐿)‘𝑁) = (seq𝑀( + , 𝐺)‘𝑁))

Theoremseq3f1o 10118* Rearrange a sum via an arbitrary bijection on (𝑀...𝑁). (Contributed by Mario Carneiro, 27-Feb-2014.) (Revised by Jim Kingdon, 3-Nov-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑦 + 𝑥))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆𝑧𝑆)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑𝐹:(𝑀...𝑁)–1-1-onto→(𝑀...𝑁))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐻𝑥) ∈ 𝑆)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐻𝑘) = (𝐺‘(𝐹𝑘)))       (𝜑 → (seq𝑀( + , 𝐻)‘𝑁) = (seq𝑀( + , 𝐺)‘𝑁))

Theoremser3add 10119* The sum of two infinite series. (Contributed by NM, 17-Mar-2005.) (Revised by Jim Kingdon, 4-Oct-2022.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐻𝑘) = ((𝐹𝑘) + (𝐺𝑘)))       (𝜑 → (seq𝑀( + , 𝐻)‘𝑁) = ((seq𝑀( + , 𝐹)‘𝑁) + (seq𝑀( + , 𝐺)‘𝑁)))

Theoremser3sub 10120* The difference of two infinite series. (Contributed by NM, 17-Mar-2005.) (Revised by Jim Kingdon, 22-Apr-2023.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℂ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐻𝑘) = ((𝐹𝑘) − (𝐺𝑘)))       (𝜑 → (seq𝑀( + , 𝐻)‘𝑁) = ((seq𝑀( + , 𝐹)‘𝑁) − (seq𝑀( + , 𝐺)‘𝑁)))

Theoremseq3id3 10121* A sequence that consists entirely of "zeroes" sums to "zero". More precisely, a constant sequence with value an element which is a + -idempotent sums (or "+'s") to that element. (Contributed by Mario Carneiro, 15-Dec-2014.) (Revised by Jim Kingdon, 8-Apr-2023.)
(𝜑 → (𝑍 + 𝑍) = 𝑍)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑥 ∈ (𝑀...𝑁)) → (𝐹𝑥) = 𝑍)    &   (𝜑𝑍𝑆)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)       (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) = 𝑍)

Theoremseq3id 10122* Discarding the first few terms of a sequence that starts with all zeroes (or any element which is a left-identity for +) has no effect on its sum. (Contributed by Mario Carneiro, 13-Jul-2013.) (Revised by Jim Kingdon, 8-Apr-2023.)
((𝜑𝑥𝑆) → (𝑍 + 𝑥) = 𝑥)    &   (𝜑𝑍𝑆)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑 → (𝐹𝑁) ∈ 𝑆)    &   ((𝜑𝑥 ∈ (𝑀...(𝑁 − 1))) → (𝐹𝑥) = 𝑍)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)       (𝜑 → (seq𝑀( + , 𝐹) ↾ (ℤ𝑁)) = seq𝑁( + , 𝐹))

Theoremseq3id2 10123* The last few partial sums of a sequence that ends with all zeroes (or any element which is a right-identity for +) are all the same. (Contributed by Mario Carneiro, 13-Jul-2013.) (Revised by Jim Kingdon, 12-Nov-2022.)
((𝜑𝑥𝑆) → (𝑥 + 𝑍) = 𝑥)    &   (𝜑𝐾 ∈ (ℤ𝑀))    &   (𝜑𝑁 ∈ (ℤ𝐾))    &   (𝜑 → (seq𝑀( + , 𝐹)‘𝐾) ∈ 𝑆)    &   ((𝜑𝑥 ∈ ((𝐾 + 1)...𝑁)) → (𝐹𝑥) = 𝑍)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)       (𝜑 → (seq𝑀( + , 𝐹)‘𝐾) = (seq𝑀( + , 𝐹)‘𝑁))

Theoremseq3homo 10124* Apply a homomorphism to a sequence. (Contributed by Jim Kingdon, 10-Oct-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝐻‘(𝑥 + 𝑦)) = ((𝐻𝑥)𝑄(𝐻𝑦)))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐻‘(𝐹𝑥)) = (𝐺𝑥))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥𝑄𝑦) ∈ 𝑆)       (𝜑 → (𝐻‘(seq𝑀( + , 𝐹)‘𝑁)) = (seq𝑀(𝑄, 𝐺)‘𝑁))

Theoremseq3z 10125* If the operation + has an absorbing element 𝑍 (a.k.a. zero element), then any sequence containing a 𝑍 evaluates to 𝑍. (Contributed by Mario Carneiro, 27-May-2014.) (Revised by Jim Kingdon, 23-Apr-2023.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   ((𝜑𝑥𝑆) → (𝑍 + 𝑥) = 𝑍)    &   ((𝜑𝑥𝑆) → (𝑥 + 𝑍) = 𝑍)    &   (𝜑𝐾 ∈ (𝑀...𝑁))    &   (𝜑 → (𝐹𝐾) = 𝑍)       (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) = 𝑍)

Theoremseqfeq3 10126* Equality of series under different addition operations which agree on an additively closed subset. (Contributed by Stefan O'Rear, 21-Mar-2015.) (Revised by Mario Carneiro, 25-Apr-2016.)
(𝜑𝑀 ∈ ℤ)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑥𝑄𝑦))       (𝜑 → seq𝑀( + , 𝐹) = seq𝑀(𝑄, 𝐹))

Theoremseq3distr 10127* The distributive property for series. (Contributed by Jim Kingdon, 10-Oct-2022.)
((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝐶𝑇(𝑥 + 𝑦)) = ((𝐶𝑇𝑥) + (𝐶𝑇𝑦)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐺𝑥) ∈ 𝑆)    &   ((𝜑𝑥 ∈ (ℤ𝑀)) → (𝐹𝑥) = (𝐶𝑇(𝐺𝑥)))    &   ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥𝑇𝑦) ∈ 𝑆)    &   (𝜑𝐶𝑆)       (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) = (𝐶𝑇(seq𝑀( + , 𝐺)‘𝑁)))

Theoremser0 10128 The value of the partial sums in a zero-valued infinite series. (Contributed by Mario Carneiro, 31-Aug-2013.) (Revised by Mario Carneiro, 15-Dec-2014.)
𝑍 = (ℤ𝑀)       (𝑁𝑍 → (seq𝑀( + , (𝑍 × {0}))‘𝑁) = 0)

Theoremser0f 10129 A zero-valued infinite series is equal to the constant zero function. (Contributed by Mario Carneiro, 8-Feb-2014.)
𝑍 = (ℤ𝑀)       (𝑀 ∈ ℤ → seq𝑀( + , (𝑍 × {0})) = (𝑍 × {0}))

Theoremfser0const 10130* Simplifying an expression which turns out just to be a constant zero sequence. (Contributed by Jim Kingdon, 16-Sep-2022.)
𝑍 = (ℤ𝑀)       (𝑁𝑍 → (𝑛𝑍 ↦ if(𝑛𝑁, ((𝑍 × {0})‘𝑛), 0)) = (𝑍 × {0}))

Theoremser3ge0 10131* A finite sum of nonnegative terms is nonnegative. (Contributed by Mario Carneiro, 8-Feb-2014.) (Revised by Mario Carneiro, 27-May-2014.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℝ)    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → 0 ≤ (𝐹𝑘))       (𝜑 → 0 ≤ (seq𝑀( + , 𝐹)‘𝑁))

Theoremser3le 10132* Comparison of partial sums of two infinite series of reals. (Contributed by NM, 27-Dec-2005.) (Revised by Jim Kingdon, 23-Apr-2023.)
(𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ∈ ℝ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐺𝑘) ∈ ℝ)    &   ((𝜑𝑘 ∈ (ℤ𝑀)) → (𝐹𝑘) ≤ (𝐺𝑘))       (𝜑 → (seq𝑀( + , 𝐹)‘𝑁) ≤ (seq𝑀( + , 𝐺)‘𝑁))

3.6.6  Integer powers

Syntaxcexp 10133 Extend class notation to include exponentiation of a complex number to an integer power.
class

Definitiondf-exp 10134* Define exponentiation to nonnegative integer powers. For example, (5↑2) = 25 (ex-exp 12542).

This definition is not meant to be used directly; instead, exp0 10138 and expp1 10141 provide the standard recursive definition. The up-arrow notation is used by Donald Knuth for iterated exponentiation (Science 194, 1235-1242, 1976) and is convenient for us since we don't have superscripts.

10-Jun-2005: The definition was extended to include zero exponents, so that 0↑0 = 1 per the convention of Definition 10-4.1 of [Gleason] p. 134 (0exp0e1 10139).

4-Jun-2014: The definition was extended to include negative integer exponents. For example, (-3↑-2) = (1 / 9) (ex-exp 12542). The case 𝑥 = 0, 𝑦 < 0 gives the value (1 / 0), so we will avoid this case in our theorems. (Contributed by Raph Levien, 20-May-2004.) (Revised by NM, 15-Oct-2004.)

↑ = (𝑥 ∈ ℂ, 𝑦 ∈ ℤ ↦ if(𝑦 = 0, 1, if(0 < 𝑦, (seq1( · , (ℕ × {𝑥}))‘𝑦), (1 / (seq1( · , (ℕ × {𝑥}))‘-𝑦)))))

Theoremexp3vallem 10135 Lemma for exp3val 10136. If we take a complex number apart from zero and raise it to a positive integer power, the result is apart from zero. (Contributed by Jim Kingdon, 7-Jun-2020.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 # 0)    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (seq1( · , (ℕ × {𝐴}))‘𝑁) # 0)

Theoremexp3val 10136 Value of exponentiation to integer powers. (Contributed by Jim Kingdon, 7-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℤ ∧ (𝐴 # 0 ∨ 0 ≤ 𝑁)) → (𝐴𝑁) = if(𝑁 = 0, 1, if(0 < 𝑁, (seq1( · , (ℕ × {𝐴}))‘𝑁), (1 / (seq1( · , (ℕ × {𝐴}))‘-𝑁)))))

Theoremexpnnval 10137 Value of exponentiation to positive integer powers. (Contributed by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → (𝐴𝑁) = (seq1( · , (ℕ × {𝐴}))‘𝑁))

Theoremexp0 10138 Value of a complex number raised to the 0th power. Note that under our definition, 0↑0 = 1, following the convention used by Gleason. Part of Definition 10-4.1 of [Gleason] p. 134. (Contributed by NM, 20-May-2004.) (Revised by Mario Carneiro, 4-Jun-2014.)
(𝐴 ∈ ℂ → (𝐴↑0) = 1)

Theorem0exp0e1 10139 0↑0 = 1 (common case). This is our convention. It follows the convention used by Gleason; see Part of Definition 10-4.1 of [Gleason] p. 134. (Contributed by David A. Wheeler, 8-Dec-2018.)
(0↑0) = 1

Theoremexp1 10140 Value of a complex number raised to the first power. (Contributed by NM, 20-Oct-2004.) (Revised by Mario Carneiro, 2-Jul-2013.)
(𝐴 ∈ ℂ → (𝐴↑1) = 𝐴)

Theoremexpp1 10141 Value of a complex number raised to a nonnegative integer power plus one. Part of Definition 10-4.1 of [Gleason] p. 134. (Contributed by NM, 20-May-2005.) (Revised by Mario Carneiro, 2-Jul-2013.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴↑(𝑁 + 1)) = ((𝐴𝑁) · 𝐴))

Theoremexpnegap0 10142 Value of a complex number raised to a negative integer power. (Contributed by Jim Kingdon, 8-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℕ0) → (𝐴↑-𝑁) = (1 / (𝐴𝑁)))

Theoremexpineg2 10143 Value of a complex number raised to a negative integer power. (Contributed by Jim Kingdon, 8-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝑁 ∈ ℂ ∧ -𝑁 ∈ ℕ0)) → (𝐴𝑁) = (1 / (𝐴↑-𝑁)))

Theoremexpn1ap0 10144 A number to the negative one power is the reciprocal. (Contributed by Jim Kingdon, 8-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0) → (𝐴↑-1) = (1 / 𝐴))

Theoremexpcllem 10145* Lemma for proving nonnegative integer exponentiation closure laws. (Contributed by NM, 14-Dec-2005.)
𝐹 ⊆ ℂ    &   ((𝑥𝐹𝑦𝐹) → (𝑥 · 𝑦) ∈ 𝐹)    &   1 ∈ 𝐹       ((𝐴𝐹𝐵 ∈ ℕ0) → (𝐴𝐵) ∈ 𝐹)

Theoremexpcl2lemap 10146* Lemma for proving integer exponentiation closure laws. (Contributed by Jim Kingdon, 8-Jun-2020.)
𝐹 ⊆ ℂ    &   ((𝑥𝐹𝑦𝐹) → (𝑥 · 𝑦) ∈ 𝐹)    &   1 ∈ 𝐹    &   ((𝑥𝐹𝑥 # 0) → (1 / 𝑥) ∈ 𝐹)       ((𝐴𝐹𝐴 # 0 ∧ 𝐵 ∈ ℤ) → (𝐴𝐵) ∈ 𝐹)

Theoremnnexpcl 10147 Closure of exponentiation of nonnegative integers. (Contributed by NM, 16-Dec-2005.)
((𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℕ)

Theoremnn0expcl 10148 Closure of exponentiation of nonnegative integers. (Contributed by NM, 14-Dec-2005.)
((𝐴 ∈ ℕ0𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℕ0)

Theoremzexpcl 10149 Closure of exponentiation of integers. (Contributed by NM, 16-Dec-2005.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℤ)

Theoremqexpcl 10150 Closure of exponentiation of rationals. (Contributed by NM, 16-Dec-2005.)
((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℚ)

Theoremreexpcl 10151 Closure of exponentiation of reals. (Contributed by NM, 14-Dec-2005.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℝ)

Theoremexpcl 10152 Closure law for nonnegative integer exponentiation. (Contributed by NM, 26-May-2005.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ∈ ℂ)

Theoremrpexpcl 10153 Closure law for exponentiation of positive reals. (Contributed by NM, 24-Feb-2008.) (Revised by Mario Carneiro, 9-Sep-2014.)
((𝐴 ∈ ℝ+𝑁 ∈ ℤ) → (𝐴𝑁) ∈ ℝ+)

Theoremreexpclzap 10154 Closure of exponentiation of reals. (Contributed by Jim Kingdon, 9-Jun-2020.)
((𝐴 ∈ ℝ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴𝑁) ∈ ℝ)

Theoremqexpclz 10155 Closure of exponentiation of rational numbers. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝐴 ∈ ℚ ∧ 𝐴 ≠ 0 ∧ 𝑁 ∈ ℤ) → (𝐴𝑁) ∈ ℚ)

Theoremm1expcl2 10156 Closure of exponentiation of negative one. (Contributed by Mario Carneiro, 18-Jun-2015.)
(𝑁 ∈ ℤ → (-1↑𝑁) ∈ {-1, 1})

Theoremm1expcl 10157 Closure of exponentiation of negative one. (Contributed by Mario Carneiro, 18-Jun-2015.)
(𝑁 ∈ ℤ → (-1↑𝑁) ∈ ℤ)

Theoremexpclzaplem 10158* Closure law for integer exponentiation. Lemma for expclzap 10159 and expap0i 10166. (Contributed by Jim Kingdon, 9-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴𝑁) ∈ {𝑧 ∈ ℂ ∣ 𝑧 # 0})

Theoremexpclzap 10159 Closure law for integer exponentiation. (Contributed by Jim Kingdon, 9-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴𝑁) ∈ ℂ)

Theoremnn0expcli 10160 Closure of exponentiation of nonnegative integers. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝑁 ∈ ℕ0       (𝐴𝑁) ∈ ℕ0

Theoremnn0sqcl 10161 The square of a nonnegative integer is a nonnegative integer. (Contributed by Stefan O'Rear, 16-Oct-2014.)
(𝐴 ∈ ℕ0 → (𝐴↑2) ∈ ℕ0)

Theoremexpm1t 10162 Exponentiation in terms of predecessor exponent. (Contributed by NM, 19-Dec-2005.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → (𝐴𝑁) = ((𝐴↑(𝑁 − 1)) · 𝐴))

Theorem1exp 10163 Value of one raised to a nonnegative integer power. (Contributed by NM, 15-Dec-2005.) (Revised by Mario Carneiro, 4-Jun-2014.)
(𝑁 ∈ ℤ → (1↑𝑁) = 1)

Theoremexpap0 10164 Positive integer exponentiation is apart from zero iff its mantissa is apart from zero. That it is easier to prove this first, and then prove expeq0 10165 in terms of it, rather than the other way around, is perhaps an illustration of the maxim "In constructive analysis, the apartness is more basic [ than ] equality." (Remark of [Geuvers], p. 1). (Contributed by Jim Kingdon, 10-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → ((𝐴𝑁) # 0 ↔ 𝐴 # 0))

Theoremexpeq0 10165 Positive integer exponentiation is 0 iff its mantissa is 0. (Contributed by NM, 23-Feb-2005.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ) → ((𝐴𝑁) = 0 ↔ 𝐴 = 0))

Theoremexpap0i 10166 Integer exponentiation is apart from zero if its mantissa is apart from zero. (Contributed by Jim Kingdon, 10-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴𝑁) # 0)

Theoremexpgt0 10167 Nonnegative integer exponentiation with a positive mantissa is positive. (Contributed by NM, 16-Dec-2005.) (Revised by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℤ ∧ 0 < 𝐴) → 0 < (𝐴𝑁))

Theoremexpnegzap 10168 Value of a complex number raised to a negative power. (Contributed by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴↑-𝑁) = (1 / (𝐴𝑁)))

Theorem0exp 10169 Value of zero raised to a positive integer power. (Contributed by NM, 19-Aug-2004.)
(𝑁 ∈ ℕ → (0↑𝑁) = 0)

Theoremexpge0 10170 Nonnegative integer exponentiation with a nonnegative mantissa is nonnegative. (Contributed by NM, 16-Dec-2005.) (Revised by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0 ∧ 0 ≤ 𝐴) → 0 ≤ (𝐴𝑁))

Theoremexpge1 10171 Nonnegative integer exponentiation with a mantissa greater than or equal to 1 is greater than or equal to 1. (Contributed by NM, 21-Feb-2005.) (Revised by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0 ∧ 1 ≤ 𝐴) → 1 ≤ (𝐴𝑁))

Theoremexpgt1 10172 Positive integer exponentiation with a mantissa greater than 1 is greater than 1. (Contributed by NM, 13-Feb-2005.) (Revised by Mario Carneiro, 4-Jun-2014.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ ∧ 1 < 𝐴) → 1 < (𝐴𝑁))

Theoremmulexp 10173 Positive integer exponentiation of a product. Proposition 10-4.2(c) of [Gleason] p. 135, restricted to nonnegative integer exponents. (Contributed by NM, 13-Feb-2005.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → ((𝐴 · 𝐵)↑𝑁) = ((𝐴𝑁) · (𝐵𝑁)))

Theoremmulexpzap 10174 Integer exponentiation of a product. (Contributed by Jim Kingdon, 10-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 # 0) ∧ 𝑁 ∈ ℤ) → ((𝐴 · 𝐵)↑𝑁) = ((𝐴𝑁) · (𝐵𝑁)))

Theoremexprecap 10175 Nonnegative integer exponentiation of a reciprocal. (Contributed by Jim Kingdon, 10-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → ((1 / 𝐴)↑𝑁) = (1 / (𝐴𝑁)))

Theoremexpadd 10176 Sum of exponents law for nonnegative integer exponentiation. Proposition 10-4.2(a) of [Gleason] p. 135. (Contributed by NM, 30-Nov-2004.)
((𝐴 ∈ ℂ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝐴↑(𝑀 + 𝑁)) = ((𝐴𝑀) · (𝐴𝑁)))

(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝑀 ∈ ℝ ∧ -𝑀 ∈ ℕ) ∧ 𝑁 ∈ ℕ0) → (𝐴↑(𝑀 + 𝑁)) = ((𝐴𝑀) · (𝐴𝑁)))

Theoremexpaddzap 10178 Sum of exponents law for integer exponentiation. (Contributed by Jim Kingdon, 10-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → (𝐴↑(𝑀 + 𝑁)) = ((𝐴𝑀) · (𝐴𝑁)))

Theoremexpmul 10179 Product of exponents law for positive integer exponentiation. Proposition 10-4.2(b) of [Gleason] p. 135, restricted to nonnegative integer exponents. (Contributed by NM, 4-Jan-2006.)
((𝐴 ∈ ℂ ∧ 𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝐴↑(𝑀 · 𝑁)) = ((𝐴𝑀)↑𝑁))

Theoremexpmulzap 10180 Product of exponents law for integer exponentiation. (Contributed by Jim Kingdon, 11-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → (𝐴↑(𝑀 · 𝑁)) = ((𝐴𝑀)↑𝑁))

Theoremm1expeven 10181 Exponentiation of negative one to an even power. (Contributed by Scott Fenton, 17-Jan-2018.)
(𝑁 ∈ ℤ → (-1↑(2 · 𝑁)) = 1)

Theoremexpsubap 10182 Exponent subtraction law for nonnegative integer exponentiation. (Contributed by Jim Kingdon, 11-Jun-2020.)
(((𝐴 ∈ ℂ ∧ 𝐴 # 0) ∧ (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → (𝐴↑(𝑀𝑁)) = ((𝐴𝑀) / (𝐴𝑁)))

Theoremexpp1zap 10183 Value of a nonzero complex number raised to an integer power plus one. (Contributed by Jim Kingdon, 11-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴↑(𝑁 + 1)) = ((𝐴𝑁) · 𝐴))

Theoremexpm1ap 10184 Value of a complex number raised to an integer power minus one. (Contributed by Jim Kingdon, 11-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (𝐴↑(𝑁 − 1)) = ((𝐴𝑁) / 𝐴))

Theoremexpdivap 10185 Nonnegative integer exponentiation of a quotient. (Contributed by Jim Kingdon, 11-Jun-2020.)
((𝐴 ∈ ℂ ∧ (𝐵 ∈ ℂ ∧ 𝐵 # 0) ∧ 𝑁 ∈ ℕ0) → ((𝐴 / 𝐵)↑𝑁) = ((𝐴𝑁) / (𝐵𝑁)))

Theoremltexp2a 10186 Ordering relationship for exponentiation. (Contributed by NM, 2-Aug-2006.) (Revised by Mario Carneiro, 4-Jun-2014.)
(((𝐴 ∈ ℝ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (1 < 𝐴𝑀 < 𝑁)) → (𝐴𝑀) < (𝐴𝑁))

Theoremleexp2a 10187 Weak ordering relationship for exponentiation. (Contributed by NM, 14-Dec-2005.) (Revised by Mario Carneiro, 5-Jun-2014.)
((𝐴 ∈ ℝ ∧ 1 ≤ 𝐴𝑁 ∈ (ℤ𝑀)) → (𝐴𝑀) ≤ (𝐴𝑁))

Theoremleexp2r 10188 Weak ordering relationship for exponentiation. (Contributed by Paul Chapman, 14-Jan-2008.) (Revised by Mario Carneiro, 29-Apr-2014.)
(((𝐴 ∈ ℝ ∧ 𝑀 ∈ ℕ0𝑁 ∈ (ℤ𝑀)) ∧ (0 ≤ 𝐴𝐴 ≤ 1)) → (𝐴𝑁) ≤ (𝐴𝑀))

Theoremleexp1a 10189 Weak mantissa ordering relationship for exponentiation. (Contributed by NM, 18-Dec-2005.)
(((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝑁 ∈ ℕ0) ∧ (0 ≤ 𝐴𝐴𝐵)) → (𝐴𝑁) ≤ (𝐵𝑁))

Theoremexple1 10190 Nonnegative integer exponentiation with a mantissa between 0 and 1 inclusive is less than or equal to 1. (Contributed by Paul Chapman, 29-Dec-2007.) (Revised by Mario Carneiro, 5-Jun-2014.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴𝐴 ≤ 1) ∧ 𝑁 ∈ ℕ0) → (𝐴𝑁) ≤ 1)

Theoremexpubnd 10191 An upper bound on 𝐴𝑁 when 2 ≤ 𝐴. (Contributed by NM, 19-Dec-2005.)
((𝐴 ∈ ℝ ∧ 𝑁 ∈ ℕ0 ∧ 2 ≤ 𝐴) → (𝐴𝑁) ≤ ((2↑𝑁) · ((𝐴 − 1)↑𝑁)))

Theoremsqval 10192 Value of the square of a complex number. (Contributed by Raph Levien, 10-Apr-2004.)
(𝐴 ∈ ℂ → (𝐴↑2) = (𝐴 · 𝐴))

Theoremsqneg 10193 The square of the negative of a number.) (Contributed by NM, 15-Jan-2006.)
(𝐴 ∈ ℂ → (-𝐴↑2) = (𝐴↑2))

Theoremsqsubswap 10194 Swap the order of subtraction in a square. (Contributed by Scott Fenton, 10-Jun-2013.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴𝐵)↑2) = ((𝐵𝐴)↑2))

Theoremsqcl 10195 Closure of square. (Contributed by NM, 10-Aug-1999.)
(𝐴 ∈ ℂ → (𝐴↑2) ∈ ℂ)

Theoremsqmul 10196 Distribution of square over multiplication. (Contributed by NM, 21-Mar-2008.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 · 𝐵)↑2) = ((𝐴↑2) · (𝐵↑2)))

Theoremsqeq0 10197 A number is zero iff its square is zero. (Contributed by NM, 11-Mar-2006.)
(𝐴 ∈ ℂ → ((𝐴↑2) = 0 ↔ 𝐴 = 0))

Theoremsqdivap 10198 Distribution of square over division. (Contributed by Jim Kingdon, 11-Jun-2020.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 # 0) → ((𝐴 / 𝐵)↑2) = ((𝐴↑2) / (𝐵↑2)))

Theoremsqne0 10199 A number is nonzero iff its square is nonzero. See also sqap0 10200 which is the same but with not equal changed to apart. (Contributed by NM, 11-Mar-2006.)
(𝐴 ∈ ℂ → ((𝐴↑2) ≠ 0 ↔ 𝐴 ≠ 0))

Theoremsqap0 10200 A number is apart from zero iff its square is apart from zero. (Contributed by Jim Kingdon, 13-Aug-2021.)
(𝐴 ∈ ℂ → ((𝐴↑2) # 0 ↔ 𝐴 # 0))

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