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Proof step 133 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 134:
nσ[n] = nσ[n1] + 1, # nσ = ∞, nσ[n] ≤ nσ[m1], n1 + m + 1 < n, nσ[n1] < nσ[m1], stutf(nσ), n ≤ m1, n1 + m + 1 < m1, n1 < m1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n] → ¬ n < m ⊦ m1 = 0, n = 0, nσ[n1 + m + 1] = nσ[n], nσ[m1] = 0, nσ[n] = nσ[m1] + 1, nσ[n1 + 1] = nσ[n1]

Rule argument

Additional information

Used simplifier rules: 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