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Proof step 42 from proof inv-get-inverse-between (previous step)

nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, n + 1 < nσ[m1], m0 < m1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < ∞.n, ∀ m. ∃ m0. m < nσ[m0], ∃ n1. m1 < n1 ∧ nσ[n1] = n + 1 ⊦ nσ[m1] = 0, m1 = 0, ∞.n = 0, nσ[n1 + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
nσ[n0] = n + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , nσ[m0] = n + 1, m0 < n0, m1 < n0, n1 + m + 1 < ∞.n, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), m0 < m1, n + 1 < nσ[m1], m0 + 1 < n0, ∀ m. ∃ m0. m < nσ[m0] ⊦ n0 = 0, nσ[n1 + m + 1] = n + 1, ∞.n = 0, m1 = 0, nσ[m1] = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1