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Proof step 163 from proof inv-get-inverse-between (previous step)

max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ⌜n1⌝ , # nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ∞, nσ[m0] = 0, nσ[n1 + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 164:
# nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n, n < nσ[m0], stutf(nσ) ⊦ max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0, nσ[n1 + m + 1] = n + 1, nσ[m0] = 0

Rule argument

Additional information

Used simplifier rules: disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic