Metamath Proof Explorer


Theorem 1wlkdlem3

Description: Lemma 3 for 1wlkd . (Contributed by AV, 22-Jan-2021)

Ref Expression
Hypotheses 1wlkd.p P = ⟨“ XY ”⟩
1wlkd.f F = ⟨“ J ”⟩
1wlkd.x φ X V
1wlkd.y φ Y V
1wlkd.l φ X = Y I J = X
1wlkd.j φ X Y X Y I J
Assertion 1wlkdlem3 φ F Word dom I

Proof

Step Hyp Ref Expression
1 1wlkd.p P = ⟨“ XY ”⟩
2 1wlkd.f F = ⟨“ J ”⟩
3 1wlkd.x φ X V
4 1wlkd.y φ Y V
5 1wlkd.l φ X = Y I J = X
6 1wlkd.j φ X Y X Y I J
7 1 2 3 4 5 6 1wlkdlem2 φ X I J
8 elfvdm X I J J dom I
9 s1cl J dom I ⟨“ J ”⟩ Word dom I
10 2 9 eqeltrid J dom I F Word dom I
11 7 8 10 3syl φ F Word dom I