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Proof step 10 from proof stutf-alltakenmax (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
stutf(nσ), ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ[m2 + 1], ∀ 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) ⊦ nσ[m2 + 1] = 0, nσ = [], ∃ m. ¬ m2 < m ∧ nσ[m] = n ∧ nσ[m + 1] = nσ[m] + 1, nσ[m2] = n ∧ nσ[m2 + 1] = nσ[m2] + 1

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0;%stutf-nonempty: ⊦ stutf(nσ) → (¬ nσ ≠ [] ↔ false);