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Proof step 111 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 112:
# nσ = ⌜m⌝ + ⌜1⌝ , n1 < m, stutf(nσ), nσ[n1] ≤ nσ[m], ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)