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Proof step 22 from proof ascsubseq-append (conclusion)

⊦ ascsubseq(ix + n0 ', x) → (¬ ascsubseq((ix + n0 ') + n ', x) ↔ ¬ precedes(n0, n, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
ascsubseq(ix + n0 ', x) ⊦ ¬ ascsubseq((ix + n0 ') + n ', x) ↔ ¬ precedes(n0, n, x)

Rule argument

Additional information

[Concretesimprules(())]