Metamath Proof Explorer


Theorem bnj986

Description: Technical lemma for bnj69 . This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj986.3 ⊢ ( 𝜒 ↔ ( 𝑛 ∈ 𝐷 ∧ 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) )
bnj986.10 ⊢ 𝐷 = ( ω ∖ { ∅ } )
bnj986.15 ⊢ ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ∧ 𝑝 = suc 𝑛 ) )
Assertion bnj986 ( 𝜒 → ∃ 𝑚 ∃ 𝑝 𝜏 )

Proof

Step Hyp Ref Expression
1 bnj986.3 ⊢ ( 𝜒 ↔ ( 𝑛 ∈ 𝐷 ∧ 𝑓 Fn 𝑛 ∧ 𝜑 ∧ 𝜓 ) )
2 bnj986.10 ⊢ 𝐷 = ( ω ∖ { ∅ } )
3 bnj986.15 ⊢ ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ∧ 𝑝 = suc 𝑛 ) )
4 2 bnj158 ⊢ ( 𝑛 ∈ 𝐷 → ∃ 𝑚 ∈ ω 𝑛 = suc 𝑚 )
5 1 4 bnj769 ⊢ ( 𝜒 → ∃ 𝑚 ∈ ω 𝑛 = suc 𝑚 )
6 5 bnj1196 ⊢ ( 𝜒 → ∃ 𝑚 ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) )
7 vex ⊢ 𝑛 ∈ V
8 7 sucex ⊢ suc 𝑛 ∈ V
9 8 isseti ⊢ ∃ 𝑝 𝑝 = suc 𝑛
10 6 9 jctir ⊢ ( 𝜒 → ( ∃ 𝑚 ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ ∃ 𝑝 𝑝 = suc 𝑛 ) )
11 exdistr ⊢ ( ∃ 𝑚 ∃ 𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) ↔ ∃ 𝑚 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ ∃ 𝑝 𝑝 = suc 𝑛 ) )
12 19.41v ⊢ ( ∃ 𝑚 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ ∃ 𝑝 𝑝 = suc 𝑛 ) ↔ ( ∃ 𝑚 ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ ∃ 𝑝 𝑝 = suc 𝑛 ) )
13 11 12 bitr2i ⊢ ( ( ∃ 𝑚 ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ ∃ 𝑝 𝑝 = suc 𝑛 ) ↔ ∃ 𝑚 ∃ 𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
14 10 13 sylib ⊢ ( 𝜒 → ∃ 𝑚 ∃ 𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
15 df-3an ⊢ ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ∧ 𝑝 = suc 𝑛 ) ↔ ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
16 3 15 bitri ⊢ ( 𝜏 ↔ ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
17 16 2exbii ⊢ ( ∃ 𝑚 ∃ 𝑝 𝜏 ↔ ∃ 𝑚 ∃ 𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
18 14 17 sylibr ⊢ ( 𝜒 → ∃ 𝑚 ∃ 𝑝 𝜏 )