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Proof step 2 from proof stutf-alltakenmax (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⌜m⌝ < # nσ, n < nσ[m], stutf(nσ) ⊦ ∃ m0. m0 < m ∧ nσ[m0] = n ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic