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Metamath Proof Explorer

Theorem prodrblem2

Description: Lemma for prodrb . (Contributed by Scott Fenton, 4-Dec-2017)

Ref Expression
Hypotheses prodmo.1 ⊢ 𝐹 = ( 𝑘 ∈ ℤ ↦ if ( 𝑘 ∈ 𝐴 , 𝐵 , 1 ) )
prodmo.2 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
prodrb.4 ⊢ ( 𝜑 → 𝑀 ∈ ℤ )
prodrb.5 ⊢ ( 𝜑 → 𝑁 ∈ ℤ )
prodrb.6 ⊢ ( 𝜑 → 𝐴 ⊆ ( ℤ≥ ‘ 𝑀 ) )
prodrb.7 ⊢ ( 𝜑 → 𝐴 ⊆ ( ℤ≥ ‘ 𝑁 ) )
Assertion prodrblem2 ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → ( seq 𝑀 ( · , 𝐹 ) ⇝ 𝐶 ↔ seq 𝑁 ( · , 𝐹 ) ⇝ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 prodmo.1 ⊢ 𝐹 = ( 𝑘 ∈ ℤ ↦ if ( 𝑘 ∈ 𝐴 , 𝐵 , 1 ) )
2 prodmo.2 ⊢ ( ( 𝜑 ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
3 prodrb.4 ⊢ ( 𝜑 → 𝑀 ∈ ℤ )
4 prodrb.5 ⊢ ( 𝜑 → 𝑁 ∈ ℤ )
5 prodrb.6 ⊢ ( 𝜑 → 𝐴 ⊆ ( ℤ≥ ‘ 𝑀 ) )
6 prodrb.7 ⊢ ( 𝜑 → 𝐴 ⊆ ( ℤ≥ ‘ 𝑁 ) )
7 4 adantr ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → 𝑁 ∈ ℤ )
8 seqex ⊢ seq 𝑀 ( · , 𝐹 ) ∈ V
9 climres ⊢ ( ( 𝑁 ∈ ℤ ∧ seq 𝑀 ( · , 𝐹 ) ∈ V ) → ( ( seq 𝑀 ( · , 𝐹 ) ↾ ( ℤ≥ ‘ 𝑁 ) ) ⇝ 𝐶 ↔ seq 𝑀 ( · , 𝐹 ) ⇝ 𝐶 ) )
10 7 8 9 sylancl ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → ( ( seq 𝑀 ( · , 𝐹 ) ↾ ( ℤ≥ ‘ 𝑁 ) ) ⇝ 𝐶 ↔ seq 𝑀 ( · , 𝐹 ) ⇝ 𝐶 ) )
11 2 adantlr ⊢ ( ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) ∧ 𝑘 ∈ 𝐴 ) → 𝐵 ∈ ℂ )
12 simpr ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) )
13 1 11 12 prodrblem ⊢ ( ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) ∧ 𝐴 ⊆ ( ℤ≥ ‘ 𝑁 ) ) → ( seq 𝑀 ( · , 𝐹 ) ↾ ( ℤ≥ ‘ 𝑁 ) ) = seq 𝑁 ( · , 𝐹 ) )
14 6 13 mpidan ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → ( seq 𝑀 ( · , 𝐹 ) ↾ ( ℤ≥ ‘ 𝑁 ) ) = seq 𝑁 ( · , 𝐹 ) )
15 14 breq1d ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → ( ( seq 𝑀 ( · , 𝐹 ) ↾ ( ℤ≥ ‘ 𝑁 ) ) ⇝ 𝐶 ↔ seq 𝑁 ( · , 𝐹 ) ⇝ 𝐶 ) )
16 10 15 bitr3d ⊢ ( ( 𝜑 ∧ 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ) → ( seq 𝑀 ( · , 𝐹 ) ⇝ 𝐶 ↔ seq 𝑁 ( · , 𝐹 ) ⇝ 𝐶 ) )