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Proof step 112 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 113:
# 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, nσ[n1 + 1] = nσ[n1]

Rule argument

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2