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Proof step 140 from proof inv-get-inverse-between (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m + 1 ≤ n ↔ m < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → ¬ n < m; (s)% ⊦ ⌜m⌝ < ∞; (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s) Used terms: nσ, n1 + m + 1, n