Metamath Proof Explorer


Theorem oppglem

Description: Lemma for oppgbas . (Contributed by Stefan O'Rear, 26-Aug-2015)

Ref Expression
Hypotheses oppgbas.1 𝑂 = ( oppg𝑅 )
oppglem.2 𝐸 = Slot 𝑁
oppglem.3 𝑁 ∈ ℕ
oppglem.4 𝑁 ≠ 2
Assertion oppglem ( 𝐸𝑅 ) = ( 𝐸𝑂 )

Proof

Step Hyp Ref Expression
1 oppgbas.1 𝑂 = ( oppg𝑅 )
2 oppglem.2 𝐸 = Slot 𝑁
3 oppglem.3 𝑁 ∈ ℕ
4 oppglem.4 𝑁 ≠ 2
5 2 3 ndxid 𝐸 = Slot ( 𝐸 ‘ ndx )
6 2 3 ndxarg ( 𝐸 ‘ ndx ) = 𝑁
7 plusgndx ( +g ‘ ndx ) = 2
8 6 7 neeq12i ( ( 𝐸 ‘ ndx ) ≠ ( +g ‘ ndx ) ↔ 𝑁 ≠ 2 )
9 4 8 mpbir ( 𝐸 ‘ ndx ) ≠ ( +g ‘ ndx )
10 5 9 setsnid ( 𝐸𝑅 ) = ( 𝐸 ‘ ( 𝑅 sSet ⟨ ( +g ‘ ndx ) , tpos ( +g𝑅 ) ⟩ ) )
11 eqid ( +g𝑅 ) = ( +g𝑅 )
12 11 1 oppgval 𝑂 = ( 𝑅 sSet ⟨ ( +g ‘ ndx ) , tpos ( +g𝑅 ) ⟩ )
13 12 fveq2i ( 𝐸𝑂 ) = ( 𝐸 ‘ ( 𝑅 sSet ⟨ ( +g ‘ ndx ) , tpos ( +g𝑅 ) ⟩ ) )
14 10 13 eqtr4i ( 𝐸𝑅 ) = ( 𝐸𝑂 )