Metamath Proof Explorer


Theorem lt2addrd

Description: If the right-hand side of a 'less than' relationship is an addition, then we can express the left-hand side as an addition, too, where each term is respectively less than each term of the original right side. (Contributed by Thierry Arnoux, 15-Mar-2017)

Ref Expression
Hypotheses lt2addrd.1 φA
lt2addrd.2 φB
lt2addrd.3 φC
lt2addrd.4 φA<B+C
Assertion lt2addrd φbcA=b+cb<Bc<C

Proof

Step Hyp Ref Expression
1 lt2addrd.1 φA
2 lt2addrd.2 φB
3 lt2addrd.3 φC
4 lt2addrd.4 φA<B+C
5 2 3 readdcld φB+C
6 5 1 resubcld φB+C-A
7 6 rehalfcld φB+C-A2
8 2 7 resubcld φBB+C-A2
9 3 7 resubcld φCB+C-A2
10 3 recnd φC
11 2 recnd φB
12 11 10 addcld φB+C
13 1 recnd φA
14 12 13 subcld φB+C-A
15 14 halfcld φB+C-A2
16 10 15 15 subsub4d φC-B+C-A2-B+C-A2=CB+C-A2+B+C-A2
17 16 oveq2d φB+CB+C-A2-B+C-A2=B+C-B+C-A2+B+C-A2
18 10 15 subcld φCB+C-A2
19 11 15 18 subadd23d φBB+C-A2+C-B+C-A2=B+CB+C-A2-B+C-A2
20 14 2halvesd φB+C-A2+B+C-A2=B+C-A
21 20 14 eqeltrd φB+C-A2+B+C-A2
22 11 10 21 addsubassd φB+C-B+C-A2+B+C-A2=B+C-B+C-A2+B+C-A2
23 17 19 22 3eqtr4d φBB+C-A2+C-B+C-A2=B+C-B+C-A2+B+C-A2
24 20 oveq2d φB+C-B+C-A2+B+C-A2=B+C-B+C-A
25 12 13 nncand φB+C-B+C-A=A
26 23 24 25 3eqtrrd φA=BB+C-A2+C-B+C-A2
27 difrp AB+CA<B+CB+C-A+
28 1 5 27 syl2anc φA<B+CB+C-A+
29 4 28 mpbid φB+C-A+
30 29 rphalfcld φB+C-A2+
31 2 30 ltsubrpd φBB+C-A2<B
32 3 30 ltsubrpd φCB+C-A2<C
33 oveq1 b=BB+C-A2b+c=B-B+C-A2+c
34 33 eqeq2d b=BB+C-A2A=b+cA=B-B+C-A2+c
35 breq1 b=BB+C-A2b<BBB+C-A2<B
36 34 35 3anbi12d b=BB+C-A2A=b+cb<Bc<CA=B-B+C-A2+cBB+C-A2<Bc<C
37 oveq2 c=CB+C-A2B-B+C-A2+c=BB+C-A2+C-B+C-A2
38 37 eqeq2d c=CB+C-A2A=B-B+C-A2+cA=BB+C-A2+C-B+C-A2
39 breq1 c=CB+C-A2c<CCB+C-A2<C
40 38 39 3anbi13d c=CB+C-A2A=B-B+C-A2+cBB+C-A2<Bc<CA=BB+C-A2+C-B+C-A2BB+C-A2<BCB+C-A2<C
41 36 40 rspc2ev BB+C-A2CB+C-A2A=BB+C-A2+C-B+C-A2BB+C-A2<BCB+C-A2<CbcA=b+cb<Bc<C
42 8 9 26 31 32 41 syl113anc φbcA=b+cb<Bc<C