Theorem pythagtriplem6

Description: Lemma for pythagtrip . Calculate ( sqrt( C - B ) ) . (Contributed by Scott Fenton, 18-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

	Ref	Expression
Assertion	pythagtriplem6	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 − 𝐵 ) ) = ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		nnz	⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℤ )
2	1	3ad2ant3	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℤ )
3		nnz	⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℤ )
4	3	3ad2ant2	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐵 ∈ ℤ )
5	2 4	zsubcld	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 − 𝐵 ) ∈ ℤ )
6	5	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐵 ) ∈ ℤ )
7		pythagtriplem10	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ) → 0 < ( 𝐶 − 𝐵 ) )
8	7	3adant3	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 0 < ( 𝐶 − 𝐵 ) )
9		elnnz	⊢ ( ( 𝐶 − 𝐵 ) ∈ ℕ ↔ ( ( 𝐶 − 𝐵 ) ∈ ℤ ∧ 0 < ( 𝐶 − 𝐵 ) ) )
10	6 8 9	sylanbrc	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐵 ) ∈ ℕ )
11	10	nnnn0d	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐵 ) ∈ ℕ0 )
12		simp3	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℕ )
13		simp2	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐵 ∈ ℕ )
14	12 13	nnaddcld	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐵 ) ∈ ℕ )
15	14	nnzd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐵 ) ∈ ℤ )
16	15	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 + 𝐵 ) ∈ ℤ )
17		nnnn0	⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℕ0 )
18	17	3ad2ant1	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℕ0 )
19	18	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 ∈ ℕ0 )
20	11 16 19	3jca	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 − 𝐵 ) ∈ ℕ0 ∧ ( 𝐶 + 𝐵 ) ∈ ℤ ∧ 𝐴 ∈ ℕ0 ) )
21		pythagtriplem4	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 − 𝐵 ) gcd ( 𝐶 + 𝐵 ) ) = 1 )
22	21	oveq1d	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐶 − 𝐵 ) gcd ( 𝐶 + 𝐵 ) ) gcd 𝐴 ) = ( 1 gcd 𝐴 ) )
23		nnz	⊢ ( 𝐴 ∈ ℕ → 𝐴 ∈ ℤ )
24	23	3ad2ant1	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℤ )
25	24	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 𝐴 ∈ ℤ )
26		1gcd	⊢ ( 𝐴 ∈ ℤ → ( 1 gcd 𝐴 ) = 1 )
27	25 26	syl	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 1 gcd 𝐴 ) = 1 )
28	22 27	eqtrd	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐶 − 𝐵 ) gcd ( 𝐶 + 𝐵 ) ) gcd 𝐴 ) = 1 )
29	20 28	jca	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐶 − 𝐵 ) ∈ ℕ0 ∧ ( 𝐶 + 𝐵 ) ∈ ℤ ∧ 𝐴 ∈ ℕ0 ) ∧ ( ( ( 𝐶 − 𝐵 ) gcd ( 𝐶 + 𝐵 ) ) gcd 𝐴 ) = 1 ) )
30		oveq1	⊢ ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) )
31	30	3ad2ant2	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) )
32	24	zcnd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐴 ∈ ℂ )
33	32	sqcld	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐴 ↑ 2 ) ∈ ℂ )
34		nncn	⊢ ( 𝐵 ∈ ℕ → 𝐵 ∈ ℂ )
35	34	3ad2ant2	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐵 ∈ ℂ )
36	35	sqcld	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐵 ↑ 2 ) ∈ ℂ )
37	33 36	pncand	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) − ( 𝐵 ↑ 2 ) ) = ( 𝐴 ↑ 2 ) )
38	37	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) − ( 𝐵 ↑ 2 ) ) = ( 𝐴 ↑ 2 ) )
39		nncn	⊢ ( 𝐶 ∈ ℕ → 𝐶 ∈ ℂ )
40	39	3ad2ant3	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → 𝐶 ∈ ℂ )
41		subsq	⊢ ( ( 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 + 𝐵 ) · ( 𝐶 − 𝐵 ) ) )
42	40 35 41	syl2anc	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 + 𝐵 ) · ( 𝐶 − 𝐵 ) ) )
43	14	nncnd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 + 𝐵 ) ∈ ℂ )
44	5	zcnd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( 𝐶 − 𝐵 ) ∈ ℂ )
45	43 44	mulcomd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( ( 𝐶 + 𝐵 ) · ( 𝐶 − 𝐵 ) ) = ( ( 𝐶 − 𝐵 ) · ( 𝐶 + 𝐵 ) ) )
46	42 45	eqtrd	⊢ ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) → ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 − 𝐵 ) · ( 𝐶 + 𝐵 ) ) )
47	46	3ad2ant1	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 ↑ 2 ) − ( 𝐵 ↑ 2 ) ) = ( ( 𝐶 − 𝐵 ) · ( 𝐶 + 𝐵 ) ) )
48	31 38 47	3eqtr3d	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐴 ↑ 2 ) = ( ( 𝐶 − 𝐵 ) · ( 𝐶 + 𝐵 ) ) )
49		coprimeprodsq	⊢ ( ( ( ( 𝐶 − 𝐵 ) ∈ ℕ0 ∧ ( 𝐶 + 𝐵 ) ∈ ℤ ∧ 𝐴 ∈ ℕ0 ) ∧ ( ( ( 𝐶 − 𝐵 ) gcd ( 𝐶 + 𝐵 ) ) gcd 𝐴 ) = 1 ) → ( ( 𝐴 ↑ 2 ) = ( ( 𝐶 − 𝐵 ) · ( 𝐶 + 𝐵 ) ) → ( 𝐶 − 𝐵 ) = ( ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ↑ 2 ) ) )
50	29 48 49	sylc	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( 𝐶 − 𝐵 ) = ( ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ↑ 2 ) )
51	50	fveq2d	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 − 𝐵 ) ) = ( √ ‘ ( ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ↑ 2 ) ) )
52	6 25	gcdcld	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ∈ ℕ0 )
53	52	nn0red	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ∈ ℝ )
54	52	nn0ge0d	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → 0 ≤ ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) )
55	53 54	sqrtsqd	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) ↑ 2 ) ) = ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) )
56	51 55	eqtrd	⊢ ( ( ( 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ ) ∧ ( ( 𝐴 ↑ 2 ) + ( 𝐵 ↑ 2 ) ) = ( 𝐶 ↑ 2 ) ∧ ( ( 𝐴 gcd 𝐵 ) = 1 ∧ ¬ 2 ∥ 𝐴 ) ) → ( √ ‘ ( 𝐶 − 𝐵 ) ) = ( ( 𝐶 − 𝐵 ) gcd 𝐴 ) )