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Proof step 1 from proof monle-leq (conclusion)

⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
mon≤(nσ), m ≤ n, ⌜n⌝ < # nσ, nσ[n] < nσ[m] ⊦ nσ[m] = 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