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Proof step 14 from proof inv-inverts (previous step)

n1 = n2 +1, nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , leninv(nσ) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[nσ[n]] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , leninv(nσ) = ⌜n2⌝ + ⌜1⌝ , ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n2 + 1)[nσ[n]] = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ: ⊦ m +1 = m + 1; from specification nat-basic2