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Proof step 41 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
# nσ = ∞, nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, n + 1 < nσ[m1], m0 < m1, stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1

Rule argument

Additional information

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