Metamath Proof Explorer


Theorem bnj945

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
Hypothesis bnj945.1 𝐺 = ( 𝑓 ∪ { ⟨ 𝑛 , 𝐶 ⟩ } )
Assertion bnj945 ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛𝐴𝑛 ) → ( 𝐺𝐴 ) = ( 𝑓𝐴 ) )

Proof

Step Hyp Ref Expression
1 bnj945.1 𝐺 = ( 𝑓 ∪ { ⟨ 𝑛 , 𝐶 ⟩ } )
2 fndm ( 𝑓 Fn 𝑛 → dom 𝑓 = 𝑛 )
3 2 ad2antll ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) → dom 𝑓 = 𝑛 )
4 3 eleq2d ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) → ( 𝐴 ∈ dom 𝑓𝐴𝑛 ) )
5 4 pm5.32i ( ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴 ∈ dom 𝑓 ) ↔ ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴𝑛 ) )
6 1 bnj941 ( 𝐶 ∈ V → ( ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) → 𝐺 Fn 𝑝 ) )
7 6 imp ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) → 𝐺 Fn 𝑝 )
8 7 fnfund ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) → Fun 𝐺 )
9 1 bnj931 𝑓𝐺
10 8 9 jctir ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) → ( Fun 𝐺𝑓𝐺 ) )
11 10 anim1i ( ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴 ∈ dom 𝑓 ) → ( ( Fun 𝐺𝑓𝐺 ) ∧ 𝐴 ∈ dom 𝑓 ) )
12 5 11 sylbir ( ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴𝑛 ) → ( ( Fun 𝐺𝑓𝐺 ) ∧ 𝐴 ∈ dom 𝑓 ) )
13 df-bnj17 ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛𝐴𝑛 ) ↔ ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛 ) ∧ 𝐴𝑛 ) )
14 3ancomb ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛 ) ↔ ( 𝐶 ∈ V ∧ 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) )
15 3anass ( ( 𝐶 ∈ V ∧ 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ↔ ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) )
16 14 15 bitri ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛 ) ↔ ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) )
17 16 anbi1i ( ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛 ) ∧ 𝐴𝑛 ) ↔ ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴𝑛 ) )
18 13 17 bitri ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛𝐴𝑛 ) ↔ ( ( 𝐶 ∈ V ∧ ( 𝑝 = suc 𝑛𝑓 Fn 𝑛 ) ) ∧ 𝐴𝑛 ) )
19 df-3an ( ( Fun 𝐺𝑓𝐺𝐴 ∈ dom 𝑓 ) ↔ ( ( Fun 𝐺𝑓𝐺 ) ∧ 𝐴 ∈ dom 𝑓 ) )
20 12 18 19 3imtr4i ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛𝐴𝑛 ) → ( Fun 𝐺𝑓𝐺𝐴 ∈ dom 𝑓 ) )
21 funssfv ( ( Fun 𝐺𝑓𝐺𝐴 ∈ dom 𝑓 ) → ( 𝐺𝐴 ) = ( 𝑓𝐴 ) )
22 20 21 syl ( ( 𝐶 ∈ V ∧ 𝑓 Fn 𝑛𝑝 = suc 𝑛𝐴𝑛 ) → ( 𝐺𝐴 ) = ( 𝑓𝐴 ) )