Metamath Proof Explorer


Theorem 1pthdlem2

Description: Lemma 2 for 1pthd . (Contributed by Alexander van der Vekens, 4-Dec-2017) (Revised by AV, 22-Jan-2021)

Ref Expression
Hypotheses 1wlkd.p P = ⟨“ XY ”⟩
1wlkd.f F = ⟨“ J ”⟩
Assertion 1pthdlem2 P 0 F P 1 ..^ F =

Proof

Step Hyp Ref Expression
1 1wlkd.p P = ⟨“ XY ”⟩
2 1wlkd.f F = ⟨“ J ”⟩
3 2 fveq2i F = ⟨“ J ”⟩
4 s1len ⟨“ J ”⟩ = 1
5 3 4 eqtri F = 1
6 5 oveq2i 1 ..^ F = 1 ..^ 1
7 fzo0 1 ..^ 1 =
8 6 7 eqtri 1 ..^ F =
9 8 imaeq2i P 1 ..^ F = P
10 9 ineq2i P 0 F P 1 ..^ F = P 0 F P
11 ima0 P =
12 11 ineq2i P 0 F P = P 0 F
13 in0 P 0 F =
14 12 13 eqtri P 0 F P =
15 10 14 eqtri P 0 F P 1 ..^ F =