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Proof step 72 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ ∃ m0. nσ[m0] = nσ[m] + 1

Proof step 72 from proof inv-get-inverse-between (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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