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Proof step 55 from proof leninv-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 56:
# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m] + 1, stutf(nσ), ∀ m0, m, nσ. m1 = m0 - m ∧ # nσ = ∞ ∧ nσ[m0] = nσ[m] + 1 ∧ m < m0 ∧ stutf(nσ) → (∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1), nσ[m1 + m] = nσ[m] + 1 ∧ m1 ≠ 0 → m ≤ m0 ∧ m0 < m1 + m ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1 ⊦ ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n - m = n; from specification nat