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Proof step 4 from proof stutf-monle (previous step)

stutf(nσ) ⊦ ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] ≤ nσ[n + 1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), nσ[n + 1] < nσ[n] ⊦ nσ[n] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat