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Proof step 62 from proof eq-leninv (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 63:
# nσ = ∞, nσ[m0] + 1 = nσ[n], nσ[m0 + 1] = nσ[n], nσ[n] < nσ[m], nσ[m0] < nσ[m], m0 < m, nσ[m0] < nσ[n], m0 < n, stutf(nσ), n < m → nσ[m] = nσ[n], ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ nσ[m] = nσ[n]

Rule argument

pos: left 10 terms: m abort: F discard: F computed: T

Additional information

Used terms: m