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Proof step 20 from proof inv-inverts-last (previous step)

# nσ = ⌜n⌝ + ⌜1⌝ , leninv(nσ) = ∞, (¬ n < n1 ∧ nσ[n1] = nσ[n]) ∧ (∀ m. ¬ n < m ∧ nσ[m] = nσ[n] → ¬ n1 < m), stutf(nσ) ⊦ n1 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
nσ[n1] = nσ[n], leninv(nσ) = ∞, # nσ = ⌜n⌝ + ⌜1⌝ , n1 < n, stutf(nσ), ∀ m. ¬ n < m ∧ nσ[m] = nσ[n] → ¬ n1 < m ⊦ n1 = n

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1