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Proof step 20 from proof inv-ls-succ (previous step)

# nσ = ∞, nσ[n0] = n + 1, n + 1 < nσ[m], n0 < m, stutf(nσ), ∞.n ≤ max∞(λ n0. nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n0 = 0

User applied rule all left

And obtained the following new goals

Proof step 21:
# nσ = ∞, nσ[n0] = n + 1, n0 < m, n + 1 < nσ[m], stutf(nσ), ∞.n ≤ max∞(λ n0. nσ[n0] = n).n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∃ n1. m < n1 ∧ nσ[n1] = n + 1 ⊦ n0 = 0

Rule argument

pos: left 7 terms: m abort: F discard: F computed: F

Additional information

Used terms: m