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Proof step 25 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), n ≤ m, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

User applied rule all left

And obtained the following new goals

Proof step 26:
nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), n ≤ m, m < n + 1 → nσ[n + 1] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Rule argument

pos: left 6 terms: n + 1 abort: F discard: F computed: F

Additional information

Used terms: n + 1