Metamath Proof Explorer


Theorem itg1addlem1

Description: Decompose a preimage, which is always a disjoint union. (Contributed by Mario Carneiro, 25-Jun-2014) (Proof shortened by Mario Carneiro, 11-Dec-2016)

Ref Expression
Hypotheses itg1addlem.1 ( 𝜑𝐹 : 𝑋𝑌 )
itg1addlem.2 ( 𝜑𝐴 ∈ Fin )
itg1addlem.3 ( ( 𝜑𝑘𝐴 ) → 𝐵 ⊆ ( 𝐹 “ { 𝑘 } ) )
itg1addlem.4 ( ( 𝜑𝑘𝐴 ) → 𝐵 ∈ dom vol )
itg1addlem.5 ( ( 𝜑𝑘𝐴 ) → ( vol ‘ 𝐵 ) ∈ ℝ )
Assertion itg1addlem1 ( 𝜑 → ( vol ‘ 𝑘𝐴 𝐵 ) = Σ 𝑘𝐴 ( vol ‘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 itg1addlem.1 ( 𝜑𝐹 : 𝑋𝑌 )
2 itg1addlem.2 ( 𝜑𝐴 ∈ Fin )
3 itg1addlem.3 ( ( 𝜑𝑘𝐴 ) → 𝐵 ⊆ ( 𝐹 “ { 𝑘 } ) )
4 itg1addlem.4 ( ( 𝜑𝑘𝐴 ) → 𝐵 ∈ dom vol )
5 itg1addlem.5 ( ( 𝜑𝑘𝐴 ) → ( vol ‘ 𝐵 ) ∈ ℝ )
6 4 5 jca ( ( 𝜑𝑘𝐴 ) → ( 𝐵 ∈ dom vol ∧ ( vol ‘ 𝐵 ) ∈ ℝ ) )
7 6 ralrimiva ( 𝜑 → ∀ 𝑘𝐴 ( 𝐵 ∈ dom vol ∧ ( vol ‘ 𝐵 ) ∈ ℝ ) )
8 3 adantrr ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → 𝐵 ⊆ ( 𝐹 “ { 𝑘 } ) )
9 simprr ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → 𝑥𝐵 )
10 8 9 sseldd ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → 𝑥 ∈ ( 𝐹 “ { 𝑘 } ) )
11 1 ffnd ( 𝜑𝐹 Fn 𝑋 )
12 11 adantr ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → 𝐹 Fn 𝑋 )
13 fniniseg ( 𝐹 Fn 𝑋 → ( 𝑥 ∈ ( 𝐹 “ { 𝑘 } ) ↔ ( 𝑥𝑋 ∧ ( 𝐹𝑥 ) = 𝑘 ) ) )
14 12 13 syl ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → ( 𝑥 ∈ ( 𝐹 “ { 𝑘 } ) ↔ ( 𝑥𝑋 ∧ ( 𝐹𝑥 ) = 𝑘 ) ) )
15 10 14 mpbid ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → ( 𝑥𝑋 ∧ ( 𝐹𝑥 ) = 𝑘 ) )
16 15 simprd ( ( 𝜑 ∧ ( 𝑘𝐴𝑥𝐵 ) ) → ( 𝐹𝑥 ) = 𝑘 )
17 16 ralrimivva ( 𝜑 → ∀ 𝑘𝐴𝑥𝐵 ( 𝐹𝑥 ) = 𝑘 )
18 invdisj ( ∀ 𝑘𝐴𝑥𝐵 ( 𝐹𝑥 ) = 𝑘Disj 𝑘𝐴 𝐵 )
19 17 18 syl ( 𝜑Disj 𝑘𝐴 𝐵 )
20 volfiniun ( ( 𝐴 ∈ Fin ∧ ∀ 𝑘𝐴 ( 𝐵 ∈ dom vol ∧ ( vol ‘ 𝐵 ) ∈ ℝ ) ∧ Disj 𝑘𝐴 𝐵 ) → ( vol ‘ 𝑘𝐴 𝐵 ) = Σ 𝑘𝐴 ( vol ‘ 𝐵 ) )
21 2 7 19 20 syl3anc ( 𝜑 → ( vol ‘ 𝑘𝐴 𝐵 ) = Σ 𝑘𝐴 ( vol ‘ 𝐵 ) )