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Proof step 70 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, nσ[m] = n + 1, ⌜n0⌝ < # nσ, m < n0, m < m0, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ⌜m⌝ < # nσ, stutf(nσ), ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = n + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 71:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]
Proof step 72:
mon≤(nσ) ∧ m ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m] ≤ nσ[m0], nσ[n0] = n + 1, nσ[m] = n + 1, ⌜n0⌝ < # nσ, m < n0, m < m0, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ⌜m⌝ < # nσ, stutf(nσ), ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = n + 1

Rule argument

spec: inst: name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m, m0] flags: current, []

Additional information

Used terms: nσ, m, m0