Thus, depending on whether x∗ ∈ G1

Thus, depending on whether x∗ ∈ G1 ∪ H or x∗ ∈ G2 ∪ H, we
have d(G1 ∪ G2,H) ≥ d(G1,H) or d(G1 ∪ G2,H) ≥ d(G2,H),
from which it follows that d(G1 ∪ G2,H) ≥ min{d(G1,H),
d(G2,H)}.

Results (Indonesian) 2:

Jadi, tergantung pada apakah x * ∈ G1 ∪ H atau x * ∈ G2 ∪ H, kami
telah d (G1 ∪ G2, H) ≥ d (G1, H) atau d (G1 ∪ G2, H) ≥ d (G2, H),
dari yang berikut bahwa d (G1 ∪ G2, H) ≥ min {d (G1, H),
d (G2, H)}.