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Proof step 94 from proof leninv-succ (previous step)

n0 = m1 + 1 ∧ n0 ≠ 0, # nσ = ⌜n0⌝ , stutf(nσ) ⊦ (∃ m. m < n0 ∧ nσ[m] = n) ∧ (∀ m. (m ≠ 0 → (∃ m0. m0 < n0 ∧ nσ[m0] = m - 1)) → ¬ n + 1 < m) ↔ nσ[m1] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 95:
# nσ = ⌜m1⌝ + ⌜1⌝ , stutf(nσ) ⊦ (∃ m. ¬ m1 < m ∧ nσ[m] = n) ∧ (∀ m. (m ≠ 0 → (∃ m0. ¬ m1 < m0 ∧ nσ[m0] = m - 1)) → ¬ n + 1 < m) ↔ nσ[m1] = n

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; 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-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf