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Proof step 45 from proof inv-inverts-leq (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 46:
# nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦
Proof step 54:
# nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. nσ[n0] ≠ nσ[n] ⊦

Rule argument

pos: left 6

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