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Proof step 44 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[m0] = n + 1, nσ[n0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , m0 < n0, m0 < m1, n + 1 < nσ[m1], n1 + m + 1 < ∞.n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), m1 < n0, m0 + 1 < n0, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 45:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

Rule argument

spec: <toplevel> name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m1, n0]

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m < n → m ≤ n; (s)% ⊦ ⌜m⌝ < ∞; (s)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m1, n0