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Proof step 68 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, m < n0, ⌜n0⌝ < # nσ, m < m0, ⌜m⌝ < # nσ, n < nσ[m], stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ nσ[m0] = nσ[m]

User applied rule insert lemma

And obtained the following new goals

Proof step 69:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 70:
nσ[n0] = n + 1, n < nσ[m], ⌜m⌝ < # nσ, m < m0, ⌜n0⌝ < # nσ, m < n0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, mon≤(nσ) ∧ m ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m] ≤ nσ[n0], ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = nσ[m]

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