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 ↔ ( 𝑛𝐷𝑓 Fn 𝑛𝜑𝜓 ) )
bnj986.10 𝐷 = ( ω ∖ { ∅ } )
bnj986.15 ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛 ) )
Assertion bnj986 ( 𝜒 → ∃ 𝑚𝑝 𝜏 )

Proof

Step Hyp Ref Expression
1 bnj986.3 ( 𝜒 ↔ ( 𝑛𝐷𝑓 Fn 𝑛𝜑𝜓 ) )
2 bnj986.10 𝐷 = ( ω ∖ { ∅ } )
3 bnj986.15 ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛 ) )
4 2 bnj158 ( 𝑛𝐷 → ∃ 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 ( ∃ 𝑚𝑝 ( ( 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 ) ↔ ∃ 𝑚𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
14 10 13 sylib ( 𝜒 → ∃ 𝑚𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
15 df-3an ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = 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 ( ∃ 𝑚𝑝 𝜏 ↔ ∃ 𝑚𝑝 ( ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚 ) ∧ 𝑝 = suc 𝑛 ) )
18 14 17 sylibr ( 𝜒 → ∃ 𝑚𝑝 𝜏 )