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Proof step 16 from proof stutf-alltakenmax (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
nσ[m + 1] = nσ[m] + 1, nσ[m2 + 1] = nσ[m2], m < m2, stutf(nσ), ⌜m2⌝ + ⌜1⌝ < # 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] = 0, nσ = [], ∃ m0. ¬ m2 < m0 ∧ nσ[m0] = nσ[m] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1