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Proof step 56 from proof inv-get-inverse-between (previous step)

mon≤(nσ) ∧ n2 -1 ≤ n2 ∧ ⌜n2⌝ < # nσ → nσ[n2 -1] ≤ nσ[n2], # nσ = ∞, nσ[n2] = nσ[n1] + 1, ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < n2, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n2 = 0, nσ[n1 + m + 1] = nσ[n2]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
nσ[n2] = nσ[n1] + 1, # nσ = ∞, n1 + m + 1 < n2, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[n2 - 1] ≤ nσ[n2], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n2] → ¬ n2 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n2], n2 = 0

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%ls-03: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%pred-03: ⊦ n -1 = n - 1; from specification nat