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Proof step 19 from proof stutf-alltakenmax (previous step)

nσ[m2 + 1] = nσ[m2] + 1, n < nσ[m2 + 1], ⌜m2⌝ + ⌜1⌝ < # nσ, stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1, ∀ m0, nσ, n. m0 < m2 +1 → ⌜m0⌝ < # nσ ∧ n < nσ[m0] ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ[m1] = n ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ nσ = [], ∃ 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 20:
nσ[m2 + 1] = nσ[m2] + 1, ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ[m2], stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1, ∀ m0, nσ, n. m0 < m2 +1 → ⌜m0⌝ < # nσ ∧ n < nσ[m0] ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ[m1] = n ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ nσ[m2] = n, nσ = [], ∃ m. ¬ m2 < m ∧ nσ[m] = n ∧ nσ[m + 1] = nσ[m] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ m ≠ n → (m < n + 1 ↔ m < n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic