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Proof step 9 from proof stutf-slowincrease (previous step)

nσ[0] = 0, nσ[m] + n0 + 1 < nσ[m + n0 + 1], ⌜m⌝ + ⌜n0⌝ + ⌜1⌝ < # nσ, ⌜m⌝ + ⌜n0⌝ < # nσ, nσ[m + n0] ≤ nσ[m] + n0, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ nσ = []

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m + n0 abort: F discard: F computed: F

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2 Used terms: m + n0