Metamath Proof Explorer


Theorem edgfid

Description: Utility theorem: index-independent form of df-edgf . (Contributed by AV, 16-Nov-2021)

Ref Expression
Assertion edgfid .ef = Slot ( .ef ‘ ndx )

Proof

Step Hyp Ref Expression
1 df-edgf .ef = Slot 1 8
2 1nn0 1 ∈ ℕ0
3 8nn 8 ∈ ℕ
4 2 3 decnncl 1 8 ∈ ℕ
5 1 4 ndxid .ef = Slot ( .ef ‘ ndx )