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Proof step 179 from proof inv-get-inverse-between (previous step)

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

Proof step 180:
# nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[n1] + 1) = ⌜n⌝ , n1 ≤ m0, stutf(nσ), nσ[n1] < nσ[m0], n1 + m + 1 < n, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[m0] = 0, nσ[n1 + m + 1] = nσ[n1] + 1, n = 0

Rule argument

Additional information

Used simplifier rules: disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1