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Proof step 9 from proof inv-get-inverse-between (previous step)

stutf(nσ), # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), max∞(λ n0. nσ[n0] = n).n + m + 1 < (# nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) ⊃ ⌜λ m0. max∞(λ n0. nσ[n0] = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ[n0] = m0).n) to leninv(nσ).n⌝ ) [n + 1 ], ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ (# nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) ⊃ ⌜λ m0. max∞(λ n0. nσ[n0] = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ[n0] = m0).n) to leninv(nσ).n⌝ ) [n + 1 ] = 0, nσ [ (# nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) ⊃ ⌜λ m0. max∞(λ n0. nσ[n0] = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ[n0] = m0).n) to leninv(nσ).n⌝ ) [n ] + m + 1 ] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), max∞(λ n0. nσ[n0] = n).n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[max∞(λ n0. nσ[n0] = n).n + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%sel-inf: ⊦ ⌜nnf⌝ [n] = nnf(n); from specification stream[natstream]