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Proof step 45 from proof inv-get-inverse-succ (previous step)

max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, ⌜n⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n) = ∞, nσ[n1 + 1] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , stutf(nσ), ⌜n⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic