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Proof step 182 from proof inv-get-inverse-between (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , (¬ m0 < n ∧ nσ[n] = nσ[n1] + 1) ∧ (∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] + 1 → ¬ n < m), n1 ≤ m0, stutf(nσ), nσ[n1] < nσ[m0], n1 + m + 1 < n, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[m0] = 0, nσ[n1 + m + 1] = nσ[n1] + 1, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 183:
nσ[n] = nσ[n1] + 1, # nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] ≤ nσ[m0], n1 + m + 1 < m0, n ≤ m0, 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[n] → ¬ n < m ⊦ n = 0, nσ[n1 + m + 1] = nσ[n], nσ[m0] = 0, m0 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat