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 ⊢ ( 𝜑 → 𝐹 : 𝑋 ⟶ 𝑌 )
itg1addlem.2 ⊢ ( 𝜑 → 𝐴 ∈ Fin )
itg1addlem.3 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ⊆ ( ◡ 𝐹 “ { 𝑘 } ) )
itg1addlem.4 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ dom vol )
itg1addlem.5 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → ( vol ‘ 𝐵 ) ∈ ℝ )
Assertion itg1addlem1 ( 𝜑 → ( vol ‘ ∪ 𝑘 ∈ 𝐴 𝐵 ) = Σ 𝑘 ∈ 𝐴 ( vol ‘ 𝐵 ) )

Proof

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