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Proof step 7 from proof stutf-alltaken (previous step)

stutf(nσ), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], ∀ nσ, n. stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = n) ⊦ ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0], ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
nσ[m0] < nσ[m0 + 1], nσ[m0] < n, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], ∀ nσ, n. stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = n) ⊦ n = 0, nσ[m0 + 1] = 0, ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf