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Proof step 15 from proof leq-leninv (previous step)

# nσ = ∞, nσ[m] + 1 < nσ[n], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n] = 0

User applied rule all left

And obtained the following new goals

Proof step 16:
# nσ = ∞, nσ[m] + 1 < nσ[n], stutf(nσ), m < n → nσ[n] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[n] = 0

Rule argument

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

Additional information

Used terms: n