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Proof step 175 from proof inv-get-inverse-between (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n1] < nσ[m0], n1 + m + 1 < ∞.n, stutf(nσ), n1 ≤ m0, stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ nσ[n1] + 1 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = nσ[n1] + 1), ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ[n0] = nσ[n1] + 1) ∨ (∀ n. m0 < n ∨ nσ[n] ≠ nσ[n1] + 1) ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 176:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[m1] = nσ[n1] + 1, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m0], stutf(nσ), n1 ≤ m0, m1 ≤ m0, nσ[m1] ≤ nσ[m0], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ[n0] = nσ[m1]) ∨ (∀ n. m0 < n ∨ nσ[n] ≠ nσ[m1]) ⊦ nσ[n1 + m + 1] = nσ[m1]

Rule argument

Additional information

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