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Proof step 17 from proof monls-exists-maxleq (previous step)

m0 = m2 +1, nσ[0] + 1 < nσ[m0 + 1], m0 ≤ m, ⌜m⌝ < # nσ, nσ[0] < nσ[m], mon<(nσ), nσ[0] < nσ[m0], nσ[0] + 1 < nσ[m], nσ[m0] < nσ[m0 + 1], nσ[m0] < nσ[m], nσ[0] < nσ[m0 + 1] ⊦ nσ[m] = 0, nσ[m0] = 0, nσ[m0 + 1] = 0, ∃ 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 18:
nσ[0] < nσ[2 + m2], nσ[m2 + 1] < nσ[m], nσ[m2 + 1] < nσ[2 + m2], nσ[0] + 1 < nσ[m], nσ[0] < nσ[m2 + 1], mon<(nσ), nσ[0] < nσ[m], ⌜m⌝ < # nσ, m2 < m, nσ[0] + 1 < nσ[2 + m2] ⊦ m = 0, nσ[2 + m2] = 0, nσ[m2 + 1] = 0, nσ[m] = 0, ∃ m0. nσ[m0] < nσ[m2 + 1] ∧ nσ[m2 + 1] ≤ nσ[m0 + 1] ∧ m0 ≤ m

Rule argument

Additional information

Used simplifier rules: trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2