Metamath Proof Explorer


Theorem smu01lem

Description: Lemma for smu01 and smu02 . (Contributed by Mario Carneiro, 19-Sep-2016)

Ref Expression
Hypotheses smu01lem.1 ⊢ φ → A ⊆ ℕ 0
smu01lem.2 ⊢ φ → B ⊆ ℕ 0
smu01lem.3 ⊢ φ ∧ k ∈ ℕ 0 ∧ n ∈ ℕ 0 → ¬ k ∈ A ∧ n − k ∈ B
Assertion smu01lem ⊢ φ → A smul B = ∅

Proof

Step Hyp Ref Expression
1 smu01lem.1 ⊢ φ → A ⊆ ℕ 0
2 smu01lem.2 ⊢ φ → B ⊆ ℕ 0
3 smu01lem.3 ⊢ φ ∧ k ∈ ℕ 0 ∧ n ∈ ℕ 0 → ¬ k ∈ A ∧ n − k ∈ B
4 smucl ⊢ A ⊆ ℕ 0 ∧ B ⊆ ℕ 0 → A smul B ⊆ ℕ 0
5 1 2 4 syl2anc ⊢ φ → A smul B ⊆ ℕ 0
6 5 sseld ⊢ φ → k ∈ A smul B → k ∈ ℕ 0
7 noel ⊢ ¬ k ∈ ∅
8 peano2nn0 ⊢ k ∈ ℕ 0 → k + 1 ∈ ℕ 0
9 fveqeq2 ⊢ x = 0 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ 0 = ∅
10 9 imbi2d ⊢ x = 0 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ 0 = ∅
11 fveqeq2 ⊢ x = k → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅
12 11 imbi2d ⊢ x = k → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅
13 fveqeq2 ⊢ x = k + 1 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
14 13 imbi2d ⊢ x = k + 1 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ x = ∅ ↔ φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
15 eqid ⊢ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 = seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1
16 1 2 15 smup0 ⊢ φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ 0 = ∅
17 oveq1 ⊢ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B = ∅ sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B
18 1 adantr ⊢ φ ∧ k ∈ ℕ 0 → A ⊆ ℕ 0
19 2 adantr ⊢ φ ∧ k ∈ ℕ 0 → B ⊆ ℕ 0
20 simpr ⊢ φ ∧ k ∈ ℕ 0 → k ∈ ℕ 0
21 18 19 15 20 smupp1 ⊢ φ ∧ k ∈ ℕ 0 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B
22 3 anassrs ⊢ φ ∧ k ∈ ℕ 0 ∧ n ∈ ℕ 0 → ¬ k ∈ A ∧ n − k ∈ B
23 22 ralrimiva ⊢ φ ∧ k ∈ ℕ 0 → ∀ n ∈ ℕ 0 ¬ k ∈ A ∧ n − k ∈ B
24 rabeq0 ⊢ n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B = ∅ ↔ ∀ n ∈ ℕ 0 ¬ k ∈ A ∧ n − k ∈ B
25 23 24 sylibr ⊢ φ ∧ k ∈ ℕ 0 → n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B = ∅
26 25 oveq2d ⊢ φ ∧ k ∈ ℕ 0 → ∅ sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B = ∅ sadd ∅
27 0ss ⊢ ∅ ⊆ ℕ 0
28 sadid1 ⊢ ∅ ⊆ ℕ 0 → ∅ sadd ∅ = ∅
29 27 28 mp1i ⊢ φ ∧ k ∈ ℕ 0 → ∅ sadd ∅ = ∅
30 26 29 eqtr2d ⊢ φ ∧ k ∈ ℕ 0 → ∅ = ∅ sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B
31 21 30 eqeq12d ⊢ φ ∧ k ∈ ℕ 0 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅ ↔ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B = ∅ sadd n ∈ ℕ 0 | k ∈ A ∧ n − k ∈ B
32 17 31 imbitrrid ⊢ φ ∧ k ∈ ℕ 0 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
33 32 expcom ⊢ k ∈ ℕ 0 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
34 33 a2d ⊢ k ∈ ℕ 0 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k = ∅ → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
35 10 12 14 14 16 34 nn0ind ⊢ k + 1 ∈ ℕ 0 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
36 8 35 syl ⊢ k ∈ ℕ 0 → φ → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
37 36 impcom ⊢ φ ∧ k ∈ ℕ 0 → seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 = ∅
38 37 eleq2d ⊢ φ ∧ k ∈ ℕ 0 → k ∈ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1 ↔ k ∈ ∅
39 7 38 mtbiri ⊢ φ ∧ k ∈ ℕ 0 → ¬ k ∈ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1
40 18 19 15 20 smuval ⊢ φ ∧ k ∈ ℕ 0 → k ∈ A smul B ↔ k ∈ seq 0 p ∈ 𝒫 ℕ 0 , m ∈ ℕ 0 ⟼ p sadd n ∈ ℕ 0 | m ∈ A ∧ n − m ∈ B n ∈ ℕ 0 ⟼ if n = 0 ∅ n − 1 ⁡ k + 1
41 39 40 mtbird ⊢ φ ∧ k ∈ ℕ 0 → ¬ k ∈ A smul B
42 41 ex ⊢ φ → k ∈ ℕ 0 → ¬ k ∈ A smul B
43 6 42 syld ⊢ φ → k ∈ A smul B → ¬ k ∈ A smul B
44 43 pm2.01d ⊢ φ → ¬ k ∈ A smul B
45 44 eq0rdv ⊢ φ → A smul B = ∅