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	⊢ − = ( -g ‘ 𝑅 )
Assertion	abvsubtri	⊢ ( ( 𝐹 ∈ 𝐴 ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝐹 ‘ ( 𝑋 − 𝑌 ) ) ≤ ( ( 𝐹 ‘ 𝑋 ) + ( 𝐹 ‘ 𝑌 ) ) )

Proof

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