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Proof step 183 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 184:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1] < nσ[m0], n1 + m + 1 < n, n1 + m + 1 < m0, stutf(nσ), n1 ≤ m0, n ≤ m0, nσ[n] ≤ nσ[m0], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n]

Rule argument

Additional information

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