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Proof step 36 from proof ascsubseq-rem (conclusion)

⊦ ascsubseq(ix0 + n ' + ix1, x) → (¬ ascsubseq(ix0 + ix1, x) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
ascsubseq(ix0 + n ' + ix1, x) ⊦ ascsubseq(ix0 + ix1, x)

Rule argument

Additional information

[Concretesimprules(())]