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Proof step 59 from proof leninv-inf (previous step)

n = m0 + 1 ∧ n ≠ 0, nσ[m] = m0, # nσ = ⌜n1⌝ + ⌜1⌝ , m ≤ n1, nσ[n1] + 1 < n, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ[m] = n0 - 1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
# nσ = ⌜n1⌝ + ⌜1⌝ , nσ[n1] < nσ[m], m ≤ n1, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ[m] = n0 - 1) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) 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%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2