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Proof step 2 from proof monls-exists-maxleq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
nσ[0] < n, ⌜m⌝ < # nσ, nσ[0] < nσ[m], mon<(nσ), n ≤ nσ[m] ⊦ ∃ m0. nσ[m0] < n ∧ n ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic