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Proof step 74 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 75:
# nσ = ∞, stutf(nσ), m < m0, nσ[m] < nσ[m0] ⊦ m0 = 0, nσ[m0] = 0, ∃ m0. nσ[m0] = nσ[m] + 1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1