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Proof step 109 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[n1] = n ∧ (∀ m. nσ[m] = n → ¬ n1 < m), n1 + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, n < nσ[m1], stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[n1 + m + 1] = n + 1, nσ[m1] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 110:
# nσ = ∞, n1 + m + 1 < max∞(λ n0. nσ[n0] = nσ[n1] + 1).n, nσ[n1] < nσ[m1], stutf(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ max∞(λ n0. nσ[n0] = nσ[n1] + 1).n = 0, nσ[n1 + m + 1] = nσ[n1] + 1, nσ[m1] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1