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Proof step 16 from proof stutf-maxinf-is-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# nσ = ∞, m0 ≤ n0, stutf(nσ), nσ[m0] ≤ nσ[n], m0 ≤ n, nσ[m0] < nσ[n0], ∀ n1. ∃ n0. n1 < n0 ∧ nσ[n0] = nσ[m0] ⊦ nσ[n0] = nσ[m0], nσ[n0] = 0

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat