Metamath Proof Explorer


Theorem bnj1001

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 bnj1001.3 ( 𝜒 ↔ ( 𝑛𝐷𝑓 Fn 𝑛𝜑𝜓 ) )
bnj1001.5 ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛 ) )
bnj1001.6 ( 𝜂 ↔ ( 𝑖𝑛𝑦 ∈ ( 𝑓𝑖 ) ) )
bnj1001.13 𝐷 = ( ω ∖ { ∅ } )
bnj1001.27 ( ( 𝜃𝜒𝜏𝜂 ) → 𝜒″ )
Assertion bnj1001 ( ( 𝜃𝜒𝜏𝜂 ) → ( 𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝 ) )

Proof

Step Hyp Ref Expression
1 bnj1001.3 ( 𝜒 ↔ ( 𝑛𝐷𝑓 Fn 𝑛𝜑𝜓 ) )
2 bnj1001.5 ( 𝜏 ↔ ( 𝑚 ∈ ω ∧ 𝑛 = suc 𝑚𝑝 = suc 𝑛 ) )
3 bnj1001.6 ( 𝜂 ↔ ( 𝑖𝑛𝑦 ∈ ( 𝑓𝑖 ) ) )
4 bnj1001.13 𝐷 = ( ω ∖ { ∅ } )
5 bnj1001.27 ( ( 𝜃𝜒𝜏𝜂 ) → 𝜒″ )
6 3 simplbi ( 𝜂𝑖𝑛 )
7 6 bnj708 ( ( 𝜃𝜒𝜏𝜂 ) → 𝑖𝑛 )
8 1 bnj1232 ( 𝜒𝑛𝐷 )
9 8 bnj706 ( ( 𝜃𝜒𝜏𝜂 ) → 𝑛𝐷 )
10 4 bnj923 ( 𝑛𝐷𝑛 ∈ ω )
11 9 10 syl ( ( 𝜃𝜒𝜏𝜂 ) → 𝑛 ∈ ω )
12 elnn ( ( 𝑖𝑛𝑛 ∈ ω ) → 𝑖 ∈ ω )
13 7 11 12 syl2anc ( ( 𝜃𝜒𝜏𝜂 ) → 𝑖 ∈ ω )
14 2 simp3bi ( 𝜏𝑝 = suc 𝑛 )
15 14 bnj707 ( ( 𝜃𝜒𝜏𝜂 ) → 𝑝 = suc 𝑛 )
16 nnord ( 𝑛 ∈ ω → Ord 𝑛 )
17 ordsucelsuc ( Ord 𝑛 → ( 𝑖𝑛 ↔ suc 𝑖 ∈ suc 𝑛 ) )
18 10 16 17 3syl ( 𝑛𝐷 → ( 𝑖𝑛 ↔ suc 𝑖 ∈ suc 𝑛 ) )
19 18 biimpa ( ( 𝑛𝐷𝑖𝑛 ) → suc 𝑖 ∈ suc 𝑛 )
20 eleq2 ( 𝑝 = suc 𝑛 → ( suc 𝑖𝑝 ↔ suc 𝑖 ∈ suc 𝑛 ) )
21 19 20 anim12i ( ( ( 𝑛𝐷𝑖𝑛 ) ∧ 𝑝 = suc 𝑛 ) → ( suc 𝑖 ∈ suc 𝑛 ∧ ( suc 𝑖𝑝 ↔ suc 𝑖 ∈ suc 𝑛 ) ) )
22 9 7 15 21 syl21anc ( ( 𝜃𝜒𝜏𝜂 ) → ( suc 𝑖 ∈ suc 𝑛 ∧ ( suc 𝑖𝑝 ↔ suc 𝑖 ∈ suc 𝑛 ) ) )
23 bianir ( ( suc 𝑖 ∈ suc 𝑛 ∧ ( suc 𝑖𝑝 ↔ suc 𝑖 ∈ suc 𝑛 ) ) → suc 𝑖𝑝 )
24 22 23 syl ( ( 𝜃𝜒𝜏𝜂 ) → suc 𝑖𝑝 )
25 5 13 24 3jca ( ( 𝜃𝜒𝜏𝜂 ) → ( 𝜒″𝑖 ∈ ω ∧ suc 𝑖𝑝 ) )