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Proof step 29 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# nσ = ∞, nσ[m2] + 1 = nσ[n], n < m0, stutf(nσ), ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = nσ[n]

Rule argument

Additional information

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