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Proof step 36 from proof leninv-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# nσ = ∞, nσ[m] = nσ[m1], nσ[m1 + 1] = nσ[m0], nσ[m0] = nσ[m1] + 1, nσ[m0] < nσ[m2], m < m2, m1 < m2, m1 < m0, m < m0, stutf(nσ), m ≤ m1, ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m1] + 1 < m0 ⊦ nσ[m2] = nσ[m0]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ ⌜m⌝ < ∞; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)