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Proof step 16 from proof leninv-succ (previous step)

mon≤(nσ) ∧ m ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ[m] ≤ nσ[m0], # nσ = ∞, m < m0, stutf(nσ), ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ nσ[m0] = nσ[m], m0 = 0, ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# nσ = ∞, stutf(nσ), m < m0, nσ[m] < nσ[m0], ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ m0 = 0, nσ[m0] = nσ[m], nσ[m0] = 0, ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat