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Proof step 151 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 152:
# nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n, n < nσ[m0], stutf(nσ) ⊦ nσ[max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)