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Proof step 4 from proof stutf-alltakenmax (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
m = m2 +1, stutf(nσ), ⌜m⌝ < # nσ, n < nσ[m], ∀ m0, nσ, n. m0 < m → stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n < nσ[m0] → nσ[m0] = 0 ∨ (∃ m1. m1 < m0 ∧ nσ[m1] = n ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ nσ[m] = 0, ∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

var: m

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0;%nat freely generated by 0, +1;;