PLFA agda exercises
det : ∀ {M M′ M″}
  → (M —→ M′)
  → (M —→ M″)
    --------
  → M′ ≡ M″
det (ξ-·₁ L—→L′)   (ξ-·₁ L—→L″)     =  cong₂ _·_ (det L—→L′ L—→L″) refl
det (ξ-·₁ L—→L′)   (ξ-·₂ VL M—→M″)  =  ⊥-elim (V¬—→ VL L—→L′)
det (ξ-·₁ L—→L′)   (β-ƛ _)          =  ⊥-elim (V¬—→ V-ƛ L—→L′)
det (ξ-·₂ VL _)    (ξ-·₁ L—→L″)     =  ⊥-elim (V¬—→ VL L—→L″)
det (ξ-·₂ _ M—→M′) (ξ-·₂ _ M—→M″)   =  cong₂ _·_ refl (det M—→M′ M—→M″)
det (ξ-·₂ _ M—→M′) (β-ƛ VM)         =  ⊥-elim (V¬—→ VM M—→M′)
det (β-ƛ _)        (ξ-·₁ L—→L″)     =  ⊥-elim (V¬—→ V-ƛ L—→L″)
det (β-ƛ VM)       (ξ-·₂ _ M—→M″)   =  ⊥-elim (V¬—→ VM M—→M″)
det (β-ƛ _)        (β-ƛ _)          =  refl
det (ξ-suc M—→M′)  (ξ-suc M—→M″)    =  cong `suc_ (det M—→M′ M—→M″)
det (ξ-case L—→L′) (ξ-case L—→L″)   =  cong₄ case_[zero⇒_|suc_⇒_]
                                         (det L—→L′ L—→L″) refl refl refl
det (ξ-case L—→L′) β-zero           =  ⊥-elim (V¬—→ V-zero L—→L′)
det (ξ-case L—→L′) (β-suc VL)       =  ⊥-elim (V¬—→ (V-suc VL) L—→L′)
det β-zero         (ξ-case M—→M″)   =  ⊥-elim (V¬—→ V-zero M—→M″)
det β-zero         β-zero           =  refl
det (β-suc VL)     (ξ-case L—→L″)   =  ⊥-elim (V¬—→ (V-suc VL) L—→L″)
det (β-suc _)      (β-suc _)        =  refl
det β-μ            β-μ              =  refl