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Proof step 134 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n] = nσ[n1] + 1, n1 < m1, n1 + m + 1 < n, nσ[n1] < nσ[m1], n1 + m + 1 < m1, stutf(nσ), nσ[n] ≤ nσ[m1], n ≤ m1, nσ[n1 + 1] = nσ[n1] → ¬ n1 < n1 + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n] → ¬ n < m ⊦ nσ[n1 + m + 1] = nσ[n], nσ[n] = nσ[m1] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 135:
# nσ = ∞, nσ[n] = nσ[n1] + 1, n1 + m + 1 < m1, nσ[n1] < nσ[m1], n1 + m + 1 < n, n1 < m1, stutf(nσ), n ≤ m1, nσ[n] ≤ nσ[m1], ∀ m. nσ[m] = nσ[n] → ¬ n < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n] = nσ[m1] + 1, nσ[n1 + m + 1] = nσ[n], nσ[n1 + 1] = nσ[n1]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)