# Metamath Proof Explorer

## Theorem abvsubtri

Description: An absolute value satisfies the triangle inequality. (Contributed by Mario Carneiro, 4-Oct-2015)

Ref Expression
Hypotheses abv0.a 𝐴 = ( AbsVal ‘ 𝑅 )
abvneg.b 𝐵 = ( Base ‘ 𝑅 )
abvsubtri.p = ( -g𝑅 )
Assertion abvsubtri ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( 𝑋 𝑌 ) ) ≤ ( ( 𝐹𝑋 ) + ( 𝐹𝑌 ) ) )

### Proof

Step Hyp Ref Expression
1 abv0.a 𝐴 = ( AbsVal ‘ 𝑅 )
2 abvneg.b 𝐵 = ( Base ‘ 𝑅 )
3 abvsubtri.p = ( -g𝑅 )
4 eqid ( +g𝑅 ) = ( +g𝑅 )
5 eqid ( invg𝑅 ) = ( invg𝑅 )
6 2 4 5 3 grpsubval ( ( 𝑋𝐵𝑌𝐵 ) → ( 𝑋 𝑌 ) = ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) )
7 6 3adant1 ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝑋 𝑌 ) = ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) )
8 7 fveq2d ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( 𝑋 𝑌 ) ) = ( 𝐹 ‘ ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) ) )
9 1 abvrcl ( 𝐹𝐴𝑅 ∈ Ring )
10 9 3ad2ant1 ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → 𝑅 ∈ Ring )
11 ringgrp ( 𝑅 ∈ Ring → 𝑅 ∈ Grp )
12 10 11 syl ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → 𝑅 ∈ Grp )
13 simp3 ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → 𝑌𝐵 )
14 2 5 grpinvcl ( ( 𝑅 ∈ Grp ∧ 𝑌𝐵 ) → ( ( invg𝑅 ) ‘ 𝑌 ) ∈ 𝐵 )
15 12 13 14 syl2anc ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( ( invg𝑅 ) ‘ 𝑌 ) ∈ 𝐵 )
16 1 2 4 abvtri ( ( 𝐹𝐴𝑋𝐵 ∧ ( ( invg𝑅 ) ‘ 𝑌 ) ∈ 𝐵 ) → ( 𝐹 ‘ ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) ) ≤ ( ( 𝐹𝑋 ) + ( 𝐹 ‘ ( ( invg𝑅 ) ‘ 𝑌 ) ) ) )
17 15 16 syld3an3 ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) ) ≤ ( ( 𝐹𝑋 ) + ( 𝐹 ‘ ( ( invg𝑅 ) ‘ 𝑌 ) ) ) )
18 1 2 5 abvneg ( ( 𝐹𝐴𝑌𝐵 ) → ( 𝐹 ‘ ( ( invg𝑅 ) ‘ 𝑌 ) ) = ( 𝐹𝑌 ) )
19 18 3adant2 ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( ( invg𝑅 ) ‘ 𝑌 ) ) = ( 𝐹𝑌 ) )
20 19 oveq2d ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( ( 𝐹𝑋 ) + ( 𝐹 ‘ ( ( invg𝑅 ) ‘ 𝑌 ) ) ) = ( ( 𝐹𝑋 ) + ( 𝐹𝑌 ) ) )
21 17 20 breqtrd ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( 𝑋 ( +g𝑅 ) ( ( invg𝑅 ) ‘ 𝑌 ) ) ) ≤ ( ( 𝐹𝑋 ) + ( 𝐹𝑌 ) ) )
22 8 21 eqbrtrd ( ( 𝐹𝐴𝑋𝐵𝑌𝐵 ) → ( 𝐹 ‘ ( 𝑋 𝑌 ) ) ≤ ( ( 𝐹𝑋 ) + ( 𝐹𝑌 ) ) )