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Proof step 1 from proof inv-get-inverse-between (conclusion)

⊦ stutf(nσ) ∧ ⌜n⌝ + ⌜1⌝ < leninv(nσ) ∧ inv(nσ)[n] + m + 1 < inv(nσ)[n + 1] → nσ[inv(nσ)[n] + m + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), inv(nσ)[n] + m + 1 < inv(nσ)[n + 1] ⊦ nσ[inv(nσ)[n] + m + 1] = n + 1, inv(nσ)[n + 1] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1