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Proof step 149 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 150:
# nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. n0 < n1 ∧ nσ[n0] = n + 1).n, n < nσ[n1 - 1], stutf(nσ) ⊦ nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic