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Proof step 29 from proof eq-leninv (previous step)

nσ[m2] + 1 = nσ[n], # nσ = ∞, n < m0, stutf(nσ), ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ m0 = 0, nσ[m0] = nσ[n]

User applied rule all left

And obtained the following new goals

Proof step 30:
nσ[m2] + 1 = nσ[n], # nσ = ∞, n < m0, stutf(nσ), m2 < n → nσ[n] = nσ[n], ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ m0 = 0, nσ[m0] = nσ[n]

Rule argument

pos: left 5 terms: n abort: F discard: F computed: F

Additional information

Used terms: n