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Proof step 28 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
nσ[m2] + 1 = nσ[n], # nσ = ∞, n < m0, stutf(nσ), ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ m0 = 0, nσ[m0] = nσ[n]

Rule argument

Additional information

Used simplifier rules: less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1