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Proof step 1 from proof min-empty (conclusion)

⊦ min([], n) = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ mon<(nσ) → (nσ = [] ⊃ min(nσ, n) = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ min(nσ, n) = (# nσ).n ;⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m) ) )
Proof step 3:
mon<([]) → ([] = [] ⊃ min([], n) = 0 ;(# [] ≠ ∞ ∧ ([])[(# []).n - 1] < n ⊃ min([], n) = (# []).n ;⌜min([], n)⌝ < # [] ∧ n ≤ ([])[min([], n)] ∧ (∀ m. ⌜m⌝ < # [] ∧ n ≤ ([])[m] → min([], n) ≤ m) ) ) ⊦ min([], n) = 0

Rule argument

spec: inst: name: min-def seq: ⊦ mon<(nσ) → (nσ = [] ⊃ min(nσ, n) = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ min(nσ, n) = (# nσ).n ; ⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m) ) ) subst: [nσ, n] → [[], n] flags: current, []

Additional information

Used terms: [], n