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Proof step 202 from proof inv-get-inverse-between (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1 + m + 1] < nσ[m0], n1 + m < m0, nσ[n1 + m + 1] < nσ[n], n1 + m < n, nσ[n1] < nσ[m0], n1 + m + 1 < n, n1 + m + 1 < m0, nσ[n1 + 1] < nσ[m0], nσ[n1 + 1] < nσ[n], stutf(nσ), nσ[n1 + 1] ≤ nσ[n1 + m + 1], n1 ≤ m0, n ≤ m0, nσ[n] ≤ nσ[m0], nσ[n1 + m + 1] ≤ nσ[n1], nσ[n1 + 1] ≤ nσ[n1], ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[n1 + m + 1] = nσ[n]

User applied rule all left

And obtained the following new goals

Proof step 203:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n] = nσ[n1] + 1, nσ[n1 + 1] < nσ[n], nσ[n1 + 1] < nσ[m0], n1 + m + 1 < m0, n1 + m + 1 < n, nσ[n1] < nσ[m0], n1 + m < n, nσ[n1 + m + 1] < nσ[n], n1 + m < m0, nσ[n1 + m + 1] < nσ[m0], stutf(nσ), nσ[n1 + 1] ≤ nσ[n1], nσ[n1 + m + 1] ≤ nσ[n1], nσ[n] ≤ nσ[m0], n ≤ m0, n1 ≤ m0, nσ[n1 + 1] ≤ nσ[n1 + m + 1], ¬ m0 < n1 + 1 ∧ nσ[n1 + 1] = nσ[n1] → ¬ n1 < n1 + 1, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n] → ¬ n < m ⊦ nσ[n1 + m + 1] = nσ[n]

Rule argument

pos: left 20 terms: n1 + 1 abort: F discard: F computed: T

Additional information

Used terms: n1 + 1