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Proof step 18 from proof mininv-inverts-inner-succ (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 19:
⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n]
Proof step 20:
mon<(nσ) ∧ n < n + 1 ∧ ⌜n + 1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜n⌝ + ⌜1⌝ < # nσ, ⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, nσ[n] < nσ[n + 1] → ¬ n < min(nσ, nσ[n] + 1) ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

Rule argument

spec: inst: name: monls-ls seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n] subst: [nσ, m, n] → [nσ, n, n + 1] flags: current, []

Additional information

Used terms: nσ, n, n + 1