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Theorem smup1 15888
 Description: Rewrite smupp1 15879 using only smul instead of the internal recursive function 𝑃. (Contributed by Mario Carneiro, 20-Sep-2016.)
Hypotheses
Ref Expression
smup1.a (𝜑𝐴 ⊆ ℕ0)
smup1.b (𝜑𝐵 ⊆ ℕ0)
smup1.n (𝜑𝑁 ∈ ℕ0)
Assertion
Ref Expression
smup1 (𝜑 → ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))
Distinct variable groups:   𝐴,𝑛   𝐵,𝑛   𝑛,𝑁   𝜑,𝑛

Proof of Theorem smup1
Dummy variables 𝑚 𝑝 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 smup1.a . . 3 (𝜑𝐴 ⊆ ℕ0)
2 smup1.b . . 3 (𝜑𝐵 ⊆ ℕ0)
3 eqid 2758 . . 3 seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1)))) = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))
4 smup1.n . . 3 (𝜑𝑁 ∈ ℕ0)
51, 2, 3, 4smupp1 15879 . 2 (𝜑 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑁 + 1)) = ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))
6 peano2nn0 11974 . . . 4 (𝑁 ∈ ℕ0 → (𝑁 + 1) ∈ ℕ0)
74, 6syl 17 . . 3 (𝜑 → (𝑁 + 1) ∈ ℕ0)
81, 2, 3, 7smupval 15887 . 2 (𝜑 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑁 + 1)) = ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵))
91, 2, 3, 4smupval 15887 . . 3 (𝜑 → (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
109oveq1d 7165 . 2 (𝜑 → ((seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑁) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))
115, 8, 103eqtr3d 2801 1 (𝜑 → ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))