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Proof step 18 from proof inv-inverts-leq (previous step)

stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = nσ[n], # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]), ∞.n < n, stutf(nσ), ⌜n⌝ < # nσ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# nσ = ∞, stutf(nσ), ∞.n < n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

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