English
Let L1 and L2 be two lift subfields between F and K inside E, ordered by inclusion. Then L1 is strictly smaller than L2 precisely when L1 is contained in L2 and their underlying carriers are different.
Русский
Пусть L1 и L2 — подполные Lift между F и K внутри E, упорядованные по включению. Тогда L1 < L2 тогда и только тогда, когда L1 ⊆ L2 и их базовые carriers различны.
LaTeX
$$$\text{For intermediate field lifts } L_1,L_2:\quad L_1 < L_2 \iff L_1 \le L_2 \ \wedge\ L_1.\operatorname{carrier} \neq L_2.\operatorname{carrier}$$$
Lean4
theorem lt_iff_le_carrier_ne : L₁ < L₂ ↔ L₁ ≤ L₂ ∧ L₁.carrier ≠ L₂.carrier := by rw [lt_iff_le_and_ne, and_congr_right];
intro h; simp_rw [Ne, eq_iff_le_carrier_eq, h, true_and]