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Proof step 3 from proof monle-leq (previous step)

∀ n, m, nσ. m1 = n - m ∧ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ ∧ nσ[n] < nσ[m] → nσ[m] = 0, m1 +1 = n - m, mon≤(nσ), m ≤ n, ⌜n⌝ < # nσ, nσ[n] < nσ[m] ⊦ nσ[m] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m1 + 1 = n - m, mon≤(nσ), m ≤ n, ⌜n⌝ < # nσ, nσ[n] < nσ[m], ∀ n, m, nσ. m1 = n - m ∧ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ ∧ nσ[n] < nσ[m] → nσ[m] = 0 ⊦ nσ[m] = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2