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Proof step 36 from proof pushx-inv (conclusion)

⊦ inv(ixs, # x) → (¬ inv(pushx(# x, ixs, x + a '), # x + 1) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
inv(ixs, # x) ⊦ inv(pushx(# x, ixs, x + a '), # x + 1)

Rule argument

Additional information

[Concretesimprules(())]