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Proof step 24 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, nσ[n] = nσ[n0], n0 < n, n0 < m, nσ[n0] < nσ[m], m < n, n0 + 1 < n, stutf(nσ), ∀ n. ∃ m. nσ[m] = n ⊦ n0 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 25:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 26:
# nσ = ∞, nσ[n] = nσ[n0], n0 + 1 < n, m < n, nσ[n0] < nσ[m], n0 < m, n0 < n, stutf(nσ), mon≤(nσ) ∧ m ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m] ≤ nσ[n0], ∀ n. ∃ m. nσ[m] = n ⊦ n0 = 0

Rule argument

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

Additional information

Used terms: nσ, m, n0