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Proof step 14 from proof leq-leninv (previous step)

# nσ = ∞, nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1), ⌜n⌝ < # nσ, stutf(nσ), n1 + 1 < nσ[n] ⊦ nσ[n] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# nσ = ∞, nσ[m] + 1 < nσ[n], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n] = 0

Rule argument

Additional information

Used simplifier rules: less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf