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Proof step 63 from proof inv-inverts-leq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
# nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < m, n ≤ m, max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n ≤ m, nσ[n] ≤ nσ[m], max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = nσ[n]).n < n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf