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Proof step 203 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 204:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1 + 1] < nσ[n1], nσ[n1 + m + 1] < nσ[m0], n1 + m < m0, nσ[n1 + m + 1] < nσ[n], n1 + m < n, nσ[n1] < nσ[m0], n1 + m + 1 < n, n1 + m + 1 < m0, nσ[n1 + 1] < nσ[m0], nσ[n1 + 1] < nσ[n], n1 < m0, nσ[n1 + 1] + 1 < nσ[m0], stutf(nσ), nσ[n1 + 1] ≤ nσ[n1 + m + 1], n ≤ m0, nσ[n] ≤ nσ[m0], nσ[n1 + m + 1] ≤ nσ[n1], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + 1] = nσ[n1], nσ[n1 + m + 1] = nσ[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ ¬ m < n → ¬ m + m0 < n; (s)% ⊦ ¬ n < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)