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Proof step 62 from proof leninv-succ (previous step)

# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m] + 1, nσ[m] = nσ[m + 1], stutf(nσ), m1 ≠ 0 → m < m2 ∧ ¬ m1 + m < m2 ∧ nσ[m] = 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 ⊦ m1 = 0, ∃ 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 63:
# nσ = ∞, nσ[m2 + 1] = nσ[m2] + 1, nσ[m] = nσ[m2], nσ[(m1 + 1) + m] = nσ[m2] + 1, nσ[m2] = nσ[m + 1], m < m2, stutf(nσ), m2 ≤ m1 + m, nσ[m1 + m] = nσ[m2] + 1 → m ≤ m0 ∧ m0 < m1 + m ∧ nσ[m2] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1 ⊦ m1 = 0, ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m2] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

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