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Proof step 137 from proof inv-get-inverse-between (previous step)

# 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 138:
# 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, 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

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)