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Proof step 24 from proof stutf-maxinf-is-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# nσ = ⌜n0⌝ , m ≤ nσ[n], stutf(nσ), n < n0, ∀ n. ∃ n0. n < n0 ∧ ⌜n0⌝ < # nσ ∧ nσ[n0] = m ⊦ n0 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf