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Proof step 18 from proof monls-exists-maxleq (previous step)

m0 = m2 +1, nσ[0] < nσ[m0], nσ[m0] < nσ[m0 + 1], nσ[m0] < nσ[m], nσ[0] + 1 < nσ[m0 + 1], nσ[0] < nσ[m], ⌜m⌝ < # nσ, nσ[0] + 1 < nσ[m], nσ[0] < nσ[m0 + 1], mon<(nσ), m0 ≤ m ⊦ ∃ m1. nσ[m1] < nσ[m0] ∧ nσ[m0] ≤ nσ[m1 + 1] ∧ m1 ≤ m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
nσ[0] < nσ[2 + m2], nσ[0] < nσ[m2 + 1], nσ[m2 + 1] < nσ[2 + m2], nσ[m2 + 1] < nσ[m], nσ[0] < nσ[m], ⌜m⌝ < # nσ, nσ[0] + 1 < nσ[m], m2 < m, nσ[0] + 1 < nσ[2 + m2], mon<(nσ) ⊦ ∃ m0. nσ[m0] < nσ[m2 + 1] ∧ nσ[m2 + 1] ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%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 + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)