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Proof step 21 from proof stutf-alltakenmax (previous step)

nσ[m + 1] = nσ[m] + 1, nσ[m2 + 1] = nσ[m2] + 1, m < m2, ⌜m2⌝ + ⌜1⌝ < # nσ, stutf(nσ), nσ[m] < nσ[m2], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1, ∀ m0, nσ, n. m0 < m2 +1 → stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n < nσ[m0] → nσ[m0] = 0 ∨ (∃ m1. m1 < m0 ∧ nσ[m1] = n ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ m2 = 0, nσ[m2] = nσ[m], nσ = [], nσ[m2] = 0, ∃ m0. ¬ m2 < m0 ∧ nσ[m0] = nσ[m] ∧ nσ[m0 + 1] = nσ[m0] + 1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: m

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: m