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Proof step 174 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%succ-01: ⊦ m + 1 ≤ 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