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Proof step 43 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , m0 < m1, n + 1 < nσ[m1], n1 + m + 1 < ∞.n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∃ n1. m1 < n1 ∧ nσ[n1] = n + 1 ⊦ nσ[n1 + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
# nσ = ∞, nσ[m0] = n + 1, nσ[n0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , m0 < n0, m0 < m1, n + 1 < nσ[m1], n1 + m + 1 < ∞.n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), m1 < n0, m0 + 1 < n0, stutf(nσ), ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic