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Proof step 25 from proof leninv-zero (previous step)

# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = nσ[n0] + 1 - 1) → nσ[n0] + 1 = 0, ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = m - 1) → m = 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = m - 1) → m = 0 ⊦ ∃ m. ¬ n0 < m ∧ nσ[m] = nσ[n0]

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 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%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat