Metamath Proof Explorer


Theorem 2wlkdlem8

Description: Lemma 8 for 2wlkd . (Contributed by AV, 14-Feb-2021)

Ref Expression
Hypotheses 2wlkd.p ⊢ P = ⟨“ ABC ”⟩
2wlkd.f ⊢ F = ⟨“ JK ”⟩
2wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V
2wlkd.n ⊢ φ → A ≠ B ∧ B ≠ C
2wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K
Assertion 2wlkdlem8 ⊢ φ → F ⁡ 0 = J ∧ F ⁡ 1 = K

Proof

Step Hyp Ref Expression
1 2wlkd.p ⊢ P = ⟨“ ABC ”⟩
2 2wlkd.f ⊢ F = ⟨“ JK ”⟩
3 2wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V
4 2wlkd.n ⊢ φ → A ≠ B ∧ B ≠ C
5 2wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K
6 1 2 3 4 5 2wlkdlem7 ⊢ φ → J ∈ V ∧ K ∈ V
7 s2fv0 ⊢ J ∈ V → ⟨“ JK ”⟩ ⁡ 0 = J
8 s2fv1 ⊢ K ∈ V → ⟨“ JK ”⟩ ⁡ 1 = K
9 7 8 anim12i ⊢ J ∈ V ∧ K ∈ V → ⟨“ JK ”⟩ ⁡ 0 = J ∧ ⟨“ JK ”⟩ ⁡ 1 = K
10 6 9 syl ⊢ φ → ⟨“ JK ”⟩ ⁡ 0 = J ∧ ⟨“ JK ”⟩ ⁡ 1 = K
11 2 fveq1i ⊢ F ⁡ 0 = ⟨“ JK ”⟩ ⁡ 0
12 11 eqeq1i ⊢ F ⁡ 0 = J ↔ ⟨“ JK ”⟩ ⁡ 0 = J
13 2 fveq1i ⊢ F ⁡ 1 = ⟨“ JK ”⟩ ⁡ 1
14 13 eqeq1i ⊢ F ⁡ 1 = K ↔ ⟨“ JK ”⟩ ⁡ 1 = K
15 12 14 anbi12i ⊢ F ⁡ 0 = J ∧ F ⁡ 1 = K ↔ ⟨“ JK ”⟩ ⁡ 0 = J ∧ ⟨“ JK ”⟩ ⁡ 1 = K
16 10 15 sylibr ⊢ φ → F ⁡ 0 = J ∧ F ⁡ 1 = K