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Proof step 5 from proof min-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⌜min(nσ, 0)⌝ < # nσ, mon<(nσ), ∀ m. ⌜m⌝ < # nσ → min(nσ, 0) ≤ m ⊦ min(nσ, 0) = 0, nσ = []

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)% ⊦ ¬ m < 0; (s)