Metamath Proof Explorer


Theorem axlowdimlem11

Description: Lemma for axlowdim . Calculate the value of Q at its distinguished point. (Contributed by Scott Fenton, 21-Apr-2013)

Ref Expression
Hypothesis axlowdimlem10.1 𝑄 = ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ∪ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) )
Assertion axlowdimlem11 ( 𝑄 ‘ ( 𝐼 + 1 ) ) = 1

Proof

Step Hyp Ref Expression
1 axlowdimlem10.1 𝑄 = ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ∪ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) )
2 1 fveq1i ( 𝑄 ‘ ( 𝐼 + 1 ) ) = ( ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ∪ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) ) ‘ ( 𝐼 + 1 ) )
3 ovex ( 𝐼 + 1 ) ∈ V
4 1ex 1 ∈ V
5 3 4 fnsn { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } Fn { ( 𝐼 + 1 ) }
6 c0ex 0 ∈ V
7 6 fconst ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) : ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ⟶ { 0 }
8 ffn ( ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) : ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ⟶ { 0 } → ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) Fn ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) )
9 7 8 ax-mp ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) Fn ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } )
10 disjdif ( { ( 𝐼 + 1 ) } ∩ ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ) = ∅
11 3 snid ( 𝐼 + 1 ) ∈ { ( 𝐼 + 1 ) }
12 10 11 pm3.2i ( ( { ( 𝐼 + 1 ) } ∩ ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ) = ∅ ∧ ( 𝐼 + 1 ) ∈ { ( 𝐼 + 1 ) } )
13 fvun1 ( ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } Fn { ( 𝐼 + 1 ) } ∧ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) Fn ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ∧ ( ( { ( 𝐼 + 1 ) } ∩ ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) ) = ∅ ∧ ( 𝐼 + 1 ) ∈ { ( 𝐼 + 1 ) } ) ) → ( ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ∪ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) ) ‘ ( 𝐼 + 1 ) ) = ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ‘ ( 𝐼 + 1 ) ) )
14 5 9 12 13 mp3an ( ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ∪ ( ( ( 1 ... 𝑁 ) ∖ { ( 𝐼 + 1 ) } ) × { 0 } ) ) ‘ ( 𝐼 + 1 ) ) = ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ‘ ( 𝐼 + 1 ) )
15 3 4 fvsn ( { ⟨ ( 𝐼 + 1 ) , 1 ⟩ } ‘ ( 𝐼 + 1 ) ) = 1
16 2 14 15 3eqtri ( 𝑄 ‘ ( 𝐼 + 1 ) ) = 1