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Proof step 158 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 159:
# nσ = ⌜m0⌝ + ⌜1⌝ , m1 ≤ m0, ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[m1] + 1).n, nσ[m1] < nσ[m0], stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ[n1] = nσ[m1]) ∨ (∀ n0. m0 < n0 ∨ nσ[n0] ≠ nσ[m1]) ⊦ max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[m1] + 1).n = 0, nσ[∞.n + m + 1] = nσ[m1] + 1, nσ[m0] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lels-03: ⊦ m < n → m ≤ n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%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