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Proof step 9 from proof stutf-alltaken (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
nσ[m0] < nσ[m0 + 1], nσ[m0] < n, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m0⌝ + ⌜1⌝ < # nσ, n ≤ nσ[m0 + 1], m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0 ⊦ n = 0, nσ[m0 + 1] = 0, ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Proof step 9 from proof stutf-alltaken (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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