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Proof step 1 from proof stutf-alltakenmax (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
stutf(nσ), ⌜m⌝ < # nσ, n < nσ[m] ⊦ nσ[m] = 0, ∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1