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Metamath Proof Explorer

Theorem rexanuz2

Description: Combine two different upper integer properties into one. (Contributed by Mario Carneiro, 26-Dec-2013)

Ref Expression
Hypothesis rexuz3.1 ⊢ 𝑍 = ( ℤ≥ ‘ 𝑀 )
Assertion rexanuz2 ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) )

Proof

Step Hyp Ref Expression
1 rexuz3.1 ⊢ 𝑍 = ( ℤ≥ ‘ 𝑀 )
2 eluzel2 ⊢ ( 𝑗 ∈ ( ℤ≥ ‘ 𝑀 ) → 𝑀 ∈ ℤ )
3 2 1 eleq2s ⊢ ( 𝑗 ∈ 𝑍 → 𝑀 ∈ ℤ )
4 3 a1d ⊢ ( 𝑗 ∈ 𝑍 → ( ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) → 𝑀 ∈ ℤ ) )
5 4 rexlimiv ⊢ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) → 𝑀 ∈ ℤ )
6 3 a1d ⊢ ( 𝑗 ∈ 𝑍 → ( ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 → 𝑀 ∈ ℤ ) )
7 6 rexlimiv ⊢ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 → 𝑀 ∈ ℤ )
8 7 adantr ⊢ ( ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) → 𝑀 ∈ ℤ )
9 1 rexuz3 ⊢ ( 𝑀 ∈ ℤ → ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ) )
10 rexanuz ⊢ ( ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ( ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) )
11 1 rexuz3 ⊢ ( 𝑀 ∈ ℤ → ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ↔ ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ) )
12 1 rexuz3 ⊢ ( 𝑀 ∈ ℤ → ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ↔ ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) )
13 11 12 anbi12d ⊢ ( 𝑀 ∈ ℤ → ( ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) ↔ ( ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) ) )
14 10 13 bitr4id ⊢ ( 𝑀 ∈ ℤ → ( ∃ 𝑗 ∈ ℤ ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) ) )
15 9 14 bitrd ⊢ ( 𝑀 ∈ ℤ → ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) ) )
16 5 8 15 pm5.21nii ⊢ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) ( 𝜑 ∧ 𝜓 ) ↔ ( ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜑 ∧ ∃ 𝑗 ∈ 𝑍 ∀ 𝑘 ∈ ( ℤ≥ ‘ 𝑗 ) 𝜓 ) )