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Proof step 15 from proof stutf-alltaken (previous step)

nσ[0] = 0, nσ[m0] < nσ[m0 + 1], nσ[m0] < n, ⌜m0⌝ < # nσ, ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0, ⌜m0⌝ + ⌜1⌝ < # nσ → nσ[m0 + 1] = nσ[m0] ∨ nσ[m0 + 1] = nσ[m0] + 1 ⊦ nσ = [], ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
nσ[0] = 0, nσ[m0 + 1] = nσ[m0] + 1, ⌜m0⌝ < # nσ, ⌜m0⌝ + ⌜1⌝ < # nσ, m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0 ⊦ nσ = [], ∃ m. m ≤ m0 + 1 ∧ nσ[m] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m ≤ m; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ m + 1 ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ ¬ n < n; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic