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Proof step 169 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 170:
# nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[n1] + 1) = ∞, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m0], stutf(nσ), n1 ≤ m0, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic