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Proof step 10 from proof monls-exists-maxleq (previous step)

nσ[0] < nσ[m], ⌜m⌝ < # nσ, n ≤ nσ[m], nσ[0] < n, mon<(nσ) ⊦ n = nσ[m], n = 0, nσ[m] = 0, ∃ m0. nσ[m0] < n ∧ n ≤ nσ[m0 + 1] ∧ m0 ≤ m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < nσ[m], ⌜m⌝ < # nσ, nσ[0] < nσ[m], mon<(nσ), nσ[0] < n, nσ[0] + 1 < nσ[m] ⊦ nσ[m] = 0, n = 0, n = nσ[m], ∃ m0. nσ[m0] < n ∧ n ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat