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Proof step 23 from proof leninv-zero (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = m - 1) → m = 0 ⊦

Rule argument

Additional information

Used simplifier rules: 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%succ: ⊦ m +1 = m + 1; from specification nat-basic2