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Proof step 110 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 111:
# nσ = ⌜m⌝ + ⌜1⌝ , n1 < m, nσ[n1] ≤ nσ[m], stutf(nσ), ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ m = 0, nσ[n1 + 1] = nσ[n1] + 1

Rule argument

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1