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Proof step 116 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 117:
# nσ = ∞, nσ[m0] = nσ[n1] + 1, stutf(nσ), nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, m0 ≤ m1, nσ[m0] ≤ nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[m0]) ∨ (∀ n. nσ[n] ≠ nσ[m0]) ⊦ nσ[m1] = 0, nσ[n1 + m + 1] = nσ[m0], ∞.n = 0

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat