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Proof step 50 from proof inv-get-inverse-last (previous step)

# nσ = ∞, nσ[m1] = nσ[n1], stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[n1] + 1, ∀ m0. nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

User applied rule all left

And obtained the following new goals

Proof step 51:
# nσ = ∞, nσ[m1] = nσ[n1], stutf(nσ), m1 < n1 → nσ[n1] = nσ[n1] + 1, ∀ m0. nσ[m0] = nσ[n1] → ¬ n1 < m0, ∀ m0. m1 < m0 → nσ[m0] = nσ[n1] + 1 ⊦ nσ[n1 + m + 1] = nσ[n1] + 1

Rule argument

pos: left 4 terms: n1 abort: F discard: F computed: F

Additional information

Used terms: n1