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Proof step 135 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n] = nσ[n1] + 1, n1 + m + 1 < m1, nσ[n1] < nσ[m1], n1 + m + 1 < n, n1 < m1, stutf(nσ), n ≤ m1, nσ[n] ≤ nσ[m1], ∀ m. nσ[m] = nσ[n] → ¬ n < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n] = nσ[m1] + 1, nσ[n1 + m + 1] = nσ[n], nσ[n1 + 1] = nσ[n1]

User applied rule insert lemma

And obtained the following new goals

Proof step 136:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 137:
# nσ = ∞, nσ[n] = nσ[n1] + 1, n1 < m1, n1 + m + 1 < n, nσ[n1] < nσ[m1], n1 + m + 1 < m1, stutf(nσ), nσ[n] ≤ nσ[m1], n ≤ m1, mon≤(nσ) ∧ n1 + 1 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ[n1 + 1] ≤ nσ[n1 + m + 1], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n] → ¬ n < m ⊦ nσ[n1 + 1] = nσ[n1], nσ[n1 + m + 1] = nσ[n], nσ[n] = nσ[m1] + 1

Rule argument

spec: <toplevel> name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, n1 + 1, n1 + m + 1]

Additional information

Used terms: nσ, n1 + 1, n1 + m + 1