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Proof step 57 from proof leninv-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
nσ[(m1 + 1) + m] = nσ[m] + 1, # nσ = ∞, stutf(nσ), nσ[m1 + m] = nσ[m] + 1 ∧ m1 ≠ 0 → m ≤ m0 ∧ m0 < m1 + m ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1, nσ[m] = nσ[m + 1] ∧ m1 ≠ 0 → m < m2 ∧ ¬ m1 + m < m2 ∧ nσ[m + 1] = nσ[m2] ∧ nσ[m2 + 1] = nσ[m2] + 1 ⊦ ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat