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Description: Lemma for ipassi . Show the inner product associative law for positive integer reciprocals. (Contributed by NM, 27-Apr-2007) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | ip1i.1 | ⊢ 𝑋 = ( BaseSet ‘ 𝑈 ) | |
| ip1i.2 | ⊢ 𝐺 = ( +𝑣 ‘ 𝑈 ) | ||
| ip1i.4 | ⊢ 𝑆 = ( ·𝑠OLD ‘ 𝑈 ) | ||
| ip1i.7 | ⊢ 𝑃 = ( ·𝑖OLD ‘ 𝑈 ) | ||
| ip1i.9 | ⊢ 𝑈 ∈ CPreHilOLD | ||
| ipasslem1.b | ⊢ 𝐵 ∈ 𝑋 | ||
| Assertion | ipasslem4 | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) = ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | ip1i.1 | ⊢ 𝑋 = ( BaseSet ‘ 𝑈 ) | |
| 2 | ip1i.2 | ⊢ 𝐺 = ( +𝑣 ‘ 𝑈 ) | |
| 3 | ip1i.4 | ⊢ 𝑆 = ( ·𝑠OLD ‘ 𝑈 ) | |
| 4 | ip1i.7 | ⊢ 𝑃 = ( ·𝑖OLD ‘ 𝑈 ) | |
| 5 | ip1i.9 | ⊢ 𝑈 ∈ CPreHilOLD | |
| 6 | ipasslem1.b | ⊢ 𝐵 ∈ 𝑋 | |
| 7 | nnrecre | ⊢ ( 𝑁 ∈ ℕ → ( 1 / 𝑁 ) ∈ ℝ ) | |
| 8 | 7 | recnd | ⊢ ( 𝑁 ∈ ℕ → ( 1 / 𝑁 ) ∈ ℂ ) |
| 9 | 5 | phnvi | ⊢ 𝑈 ∈ NrmCVec |
| 10 | 1 3 | nvscl | ⊢ ( ( 𝑈 ∈ NrmCVec ∧ ( 1 / 𝑁 ) ∈ ℂ ∧ 𝐴 ∈ 𝑋 ) → ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 ) |
| 11 | 9 10 | mp3an1 | ⊢ ( ( ( 1 / 𝑁 ) ∈ ℂ ∧ 𝐴 ∈ 𝑋 ) → ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 ) |
| 12 | 8 11 | sylan | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 ) |
| 13 | 1 4 | dipcl | ⊢ ( ( 𝑈 ∈ NrmCVec ∧ ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ∈ ℂ ) |
| 14 | 9 6 13 | mp3an13 | ⊢ ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 → ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ∈ ℂ ) |
| 15 | 12 14 | syl | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ∈ ℂ ) |
| 16 | 1 4 | dipcl | ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( 𝐴 𝑃 𝐵 ) ∈ ℂ ) |
| 17 | 9 6 16 | mp3an13 | ⊢ ( 𝐴 ∈ 𝑋 → ( 𝐴 𝑃 𝐵 ) ∈ ℂ ) |
| 18 | mulcl | ⊢ ( ( ( 1 / 𝑁 ) ∈ ℂ ∧ ( 𝐴 𝑃 𝐵 ) ∈ ℂ ) → ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ∈ ℂ ) | |
| 19 | 8 17 18 | syl2an | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ∈ ℂ ) |
| 20 | nncn | ⊢ ( 𝑁 ∈ ℕ → 𝑁 ∈ ℂ ) | |
| 21 | 20 | adantr | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → 𝑁 ∈ ℂ ) |
| 22 | nnne0 | ⊢ ( 𝑁 ∈ ℕ → 𝑁 ≠ 0 ) | |
| 23 | 22 | adantr | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → 𝑁 ≠ 0 ) |
| 24 | 20 22 | recidd | ⊢ ( 𝑁 ∈ ℕ → ( 𝑁 · ( 1 / 𝑁 ) ) = 1 ) |
| 25 | 24 | oveq1d | ⊢ ( 𝑁 ∈ ℕ → ( ( 𝑁 · ( 1 / 𝑁 ) ) · ( 𝐴 𝑃 𝐵 ) ) = ( 1 · ( 𝐴 𝑃 𝐵 ) ) ) |
| 26 | 17 | mullidd | ⊢ ( 𝐴 ∈ 𝑋 → ( 1 · ( 𝐴 𝑃 𝐵 ) ) = ( 𝐴 𝑃 𝐵 ) ) |
| 27 | 25 26 | sylan9eq | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) · ( 𝐴 𝑃 𝐵 ) ) = ( 𝐴 𝑃 𝐵 ) ) |
| 28 | 24 | oveq1d | ⊢ ( 𝑁 ∈ ℕ → ( ( 𝑁 · ( 1 / 𝑁 ) ) 𝑆 𝐴 ) = ( 1 𝑆 𝐴 ) ) |
| 29 | 1 3 | nvsid | ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ 𝑋 ) → ( 1 𝑆 𝐴 ) = 𝐴 ) |
| 30 | 9 29 | mpan | ⊢ ( 𝐴 ∈ 𝑋 → ( 1 𝑆 𝐴 ) = 𝐴 ) |
| 31 | 28 30 | sylan9eq | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) 𝑆 𝐴 ) = 𝐴 ) |
| 32 | 8 | adantr | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( 1 / 𝑁 ) ∈ ℂ ) |
| 33 | simpr | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → 𝐴 ∈ 𝑋 ) | |
| 34 | 1 3 | nvsass | ⊢ ( ( 𝑈 ∈ NrmCVec ∧ ( 𝑁 ∈ ℂ ∧ ( 1 / 𝑁 ) ∈ ℂ ∧ 𝐴 ∈ 𝑋 ) ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) 𝑆 𝐴 ) = ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) ) |
| 35 | 9 34 | mpan | ⊢ ( ( 𝑁 ∈ ℂ ∧ ( 1 / 𝑁 ) ∈ ℂ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) 𝑆 𝐴 ) = ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) ) |
| 36 | 21 32 33 35 | syl3anc | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) 𝑆 𝐴 ) = ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) ) |
| 37 | 31 36 | eqtr3d | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → 𝐴 = ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) ) |
| 38 | 37 | oveq1d | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( 𝐴 𝑃 𝐵 ) = ( ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) 𝑃 𝐵 ) ) |
| 39 | nnnn0 | ⊢ ( 𝑁 ∈ ℕ → 𝑁 ∈ ℕ0 ) | |
| 40 | 39 | adantr | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → 𝑁 ∈ ℕ0 ) |
| 41 | 1 2 3 4 5 6 | ipasslem1 | ⊢ ( ( 𝑁 ∈ ℕ0 ∧ ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ∈ 𝑋 ) → ( ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) 𝑃 𝐵 ) = ( 𝑁 · ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ) ) |
| 42 | 40 12 41 | syl2anc | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 𝑆 ( ( 1 / 𝑁 ) 𝑆 𝐴 ) ) 𝑃 𝐵 ) = ( 𝑁 · ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ) ) |
| 43 | 27 38 42 | 3eqtrd | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) · ( 𝐴 𝑃 𝐵 ) ) = ( 𝑁 · ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ) ) |
| 44 | 17 | adantl | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( 𝐴 𝑃 𝐵 ) ∈ ℂ ) |
| 45 | 21 32 44 | mulassd | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( 𝑁 · ( 1 / 𝑁 ) ) · ( 𝐴 𝑃 𝐵 ) ) = ( 𝑁 · ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ) ) |
| 46 | 43 45 | eqtr3d | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( 𝑁 · ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) ) = ( 𝑁 · ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ) ) |
| 47 | 15 19 21 23 46 | mulcanad | ⊢ ( ( 𝑁 ∈ ℕ ∧ 𝐴 ∈ 𝑋 ) → ( ( ( 1 / 𝑁 ) 𝑆 𝐴 ) 𝑃 𝐵 ) = ( ( 1 / 𝑁 ) · ( 𝐴 𝑃 𝐵 ) ) ) |