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Proof step 40 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m] = nσ[m1], nσ[m1 + 1] = nσ[m0], nσ[m0] = nσ[m1] + 1, nσ[m2] = m3 + 1, nσ[m0] < nσ[m2], m < m2, m1 < m2, m1 < m0, m < m0, stutf(nσ), m ≤ m1 ⊦ nσ[m2] = nσ[m0], nσ[m2] = 0, ∃ m1. nσ[m1] = m3 ∧ ¬(∀ m. m1 < m → nσ[m] = m3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
# nσ = ∞, nσ[m2] = m3 + 1, nσ[m0] = nσ[m1] + 1, nσ[m1 + 1] = nσ[m0], nσ[m] = nσ[m1], nσ[m1] < m3, m < m0, m1 < m0, m1 < m2, m < m2, stutf(nσ), nσ[m0] ≤ m3, m ≤ m1 ⊦ m3 + 1 = nσ[m0], ∃ m4. nσ[m4] = m3 ∧ ¬(∀ m. m4 < m → nσ[m] = nσ[m4])

Rule argument

Additional information

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