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Proof step 5 from proof mon-eq (previous step)

nσ[n] = nσ[n0], ⌜n⌝ < # nσ, ⌜n0⌝ < # nσ, mon<(nσ), mon<(nσ) ∧ n < n0 ∧ ⌜n0⌝ < # nσ → nσ[n] < nσ[n0] ⊦ n = n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
nσ[n] = nσ[n0], n0 < n, ⌜n0⌝ < # nσ, ⌜n⌝ < # nσ, mon<(nσ) ⊦ n = n0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)% ⊦ ¬ n < n; (s)