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Proof step 6 from proof monls-exists-maxleq (previous step)

m = m1 +1, nσ[0] < nσ[m], ⌜m⌝ < # nσ, mon<(nσ) ⊦ ∃ 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 7:
nσ[0] < nσ[m1 + 1], ⌜m1⌝ + ⌜1⌝ < # nσ, mon<(nσ) ⊦ ∃ m. nσ[m] < nσ[m1 + 1] ∧ nσ[m1 + 1] ≤ nσ[m + 1] ∧ m ≤ m1 + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ m +1 = m + 1; (s)