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Proof step 16 from proof monls-exists-maxleq (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 17:
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

Rule argument

var: m0

Additional information

Used simplifier rules: nat freely generated by 0, +1;;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lez: ⊦ 0 ≤ m; from specification nat%ref-1: ⊦ ¬ n < n; from specification nat-basic1