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Proof step 19 from proof inv-inverts-leq (previous step)

# nσ = ∞, stutf(nσ), ∞.n < n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

User applied rule all left

And obtained the following new goals

Proof step 20:
# nσ = ∞, stutf(nσ), ∞.n < n, ∃ m0. n < m0 ∧ nσ[n] < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

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

Additional information

Used terms: n