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Proof step 5 from proof monls-exists-maxleq (previous step)

m = m1 +1, ⌜m⌝ < # nσ, nσ[0] < nσ[m], mon<(nσ) ⊦ nσ[m] = 0, ∃ m0. nσ[m0] < nσ[m] ∧ nσ[m] ≤ nσ[m0 + 1] ∧ m0 ≤ m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
mon<(nσ), ⌜m1⌝ + ⌜1⌝ < # nσ, nσ[0] < nσ[m1 + 1] ⊦ nσ[m1 + 1] = 0, ∃ m. nσ[m] < nσ[m1 + 1] ∧ nσ[m1 + 1] ≤ nσ[m + 1] ∧ m ≤ m1 + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ: ⊦ m +1 = m + 1; from specification nat-basic2