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Proof step 10 from proof leninv-inf-strong (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
# nσ = ∞, stutf(nσ), # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → (∃ m1. m0 < nσ[m1]), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m. m0 + 1 < nσ[m]

Rule argument

pos: left 3 terms: nσ abort: F discard: T computed: T

Additional information

Used terms: nσ