Metamath Proof Explorer


Theorem dnibndlem2

Description: Lemma for dnibnd . (Contributed by Asger C. Ipsen, 4-Apr-2021)

Ref Expression
Hypotheses dnibndlem2.1 T = x x + 1 2 x
dnibndlem2.2 φ A
dnibndlem2.3 φ B
dnibndlem2.4 φ B + 1 2 = A + 1 2
Assertion dnibndlem2 φ T B T A B A

Proof

Step Hyp Ref Expression
1 dnibndlem2.1 T = x x + 1 2 x
2 dnibndlem2.2 φ A
3 dnibndlem2.3 φ B
4 dnibndlem2.4 φ B + 1 2 = A + 1 2
5 halfre 1 2
6 5 a1i φ 1 2
7 3 6 jca φ B 1 2
8 readdcl B 1 2 B + 1 2
9 7 8 syl φ B + 1 2
10 reflcl B + 1 2 B + 1 2
11 9 10 syl φ B + 1 2
12 11 recnd φ B + 1 2
13 3 recnd φ B
14 12 13 subcld φ B + 1 2 B
15 14 abscld φ B + 1 2 B
16 15 recnd φ B + 1 2 B
17 4 12 eqeltrrd φ A + 1 2
18 2 recnd φ A
19 17 18 subcld φ A + 1 2 A
20 19 abscld φ A + 1 2 A
21 20 recnd φ A + 1 2 A
22 16 21 subcld φ B + 1 2 B A + 1 2 A
23 22 abscld φ B + 1 2 B A + 1 2 A
24 14 19 subcld φ B + 1 2 - B - A + 1 2 A
25 24 abscld φ B + 1 2 - B - A + 1 2 A
26 13 18 subcld φ B A
27 26 abscld φ B A
28 14 19 abs2difabsd φ B + 1 2 B A + 1 2 A B + 1 2 - B - A + 1 2 A
29 12 18 13 nnncan1d φ B + 1 2 - A - B + 1 2 B = B A
30 29 eqcomd φ B A = B + 1 2 - A - B + 1 2 B
31 30 fveq2d φ B A = B + 1 2 - A - B + 1 2 B
32 4 oveq1d φ B + 1 2 A = A + 1 2 A
33 32 oveq1d φ B + 1 2 - A - B + 1 2 B = A + 1 2 - A - B + 1 2 B
34 33 fveq2d φ B + 1 2 - A - B + 1 2 B = A + 1 2 - A - B + 1 2 B
35 19 14 abssubd φ A + 1 2 - A - B + 1 2 B = B + 1 2 - B - A + 1 2 A
36 31 34 35 3eqtrd φ B A = B + 1 2 - B - A + 1 2 A
37 27 leidd φ B A B A
38 36 37 eqbrtrrd φ B + 1 2 - B - A + 1 2 A B A
39 23 25 27 28 38 letrd φ B + 1 2 B A + 1 2 A B A
40 1 2 3 dnibndlem1 φ T B T A B A B + 1 2 B A + 1 2 A B A
41 39 40 mpbird φ T B T A B A