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Proof step 24 from proof pushx-ordered (conclusion)

⊦ ordered(ixs, x) ∧ inv(ixs, n) ∧ n < # x → (¬ ordered(pushx(n, ixs, x), x) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
n < # x, inv(ixs, n), ordered(ixs, x) ⊦ x = [], ordered(pushx(n, ixs, x), x)

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(list-data,,l, ⊦ # x = 0 ↔ x = []) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0)))]