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Proof step 6 from proof stutf-alltakenmax (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ[m2 + 1], stutf(nσ), ∀ 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) ⊦ ∃ m. ¬ m2 < m ∧ nσ[m] = n ∧ nσ[m + 1] = nσ[m] + 1

Rule argument

Additional information

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