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Proof step 71 from proof leninv-succ (previous step)

m ≤ m0 ∧ m0 < m1 + m ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1, nσ[(m1 + 1) + m] = nσ[m] + 1, # nσ = ∞, stutf(nσ) ⊦ nσ[m] = nσ[m + 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 72:
# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m0] + 1, nσ[m0 + 1] = nσ[m0] + 1, nσ[m] = nσ[m0], stutf(nσ), m0 < m1 + m, m ≤ m0 ⊦ m1 + m = 0, nσ[m0] = nσ[m + 1], m1 = 0, ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1