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Proof step 88 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
# nσ = ∞, stutf(nσ), n < nσ[m1], max∞(λ n0. nσ[n0] = n).n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[m1] = 0, nσ[max∞(λ n0. nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. nσ[n0] = n + 1).n = 0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1