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Proof step 21 from proof inv-inverts-leq (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 22:
# nσ = ∞, ∞.n < m, nσ[n] < nσ[m], n < m, ∞.n < n, stutf(nσ), ∞.n + 1 < m, ∃ n1. m < n1 ∧ nσ[n1] = nσ[n], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Rule argument

pos: left 8 terms: m abort: F discard: T computed: F

Additional information

Used terms: m