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Proof step 16 from proof leq-leninv (previous step)

# nσ = ∞, nσ[m] + 1 < nσ[n], stutf(nσ), m < n → nσ[n] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# nσ = ∞, stutf(nσ), nσ[m] + 1 < nσ[n], n ≤ m, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n] = 0, nσ[n] = nσ[m] + 1

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1