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Proof step 7 from proof stutf-alltakenmax (previous step)

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, ¬ m2 < m2 ∧ nσ[m2] = n ∧ nσ[m2 + 1] = nσ[m2] + 1, ∃ 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 8:
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

Rule argument

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1