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Proof step 11 from proof stutf-maxinf-is-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# nσ = ∞, m0 ≤ n, nσ[m0] ≤ nσ[n], stutf(nσ), ∀ n1. ∃ n0. n1 < n0 ∧ nσ[n0] = nσ[m0] ⊦ ∃ n. ∀ n0. ¬ n0 < n → nσ[n0] = nσ[m0]

Rule argument

Additional information

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