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Proof step 49 from proof inv-get-inverse-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 50:
# nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ < leninv(nσ), nσ[n1 +1] = nσ[n1] → ¬ n1 < n1 +1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + 1] = nσ[n1] + 1

Rule argument

pos: left 4 terms: n1 +1 abort: F discard: F computed: T

Additional information

Used terms: n1 +1