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Proof step 13 from proof precedes-ascsubseq-2 (conclusion)

⊦ n0 ∈ ix ∧ n1 ∈ ix ∧ n0 ≠ n1 ∧ ascsubseq(ix, x) → precedes(n0, n1, x) ∨ precedes(n1, n0, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ascsubseq(ix, x), n1 ∈ ix, n0 ∈ ix ⊦ n0 = n1, precedes(n1, n0, x), precedes(n0, n1, x)

Rule argument

Additional information

[Concretesimprules(())]