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Proof step 23 from proof pushx-ordered (previous step)

n < # x, inv(ixs, n), ordered(ixs, x) ⊦ x = [], ordered(pushx(n, ixs, x), x)

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term:ixs

Additional information

[Concretesimprules((Csimpnamedseq(inv-base,,, ⊦ inv([], n)) Csimpnamedseq(orderedl-base,,, ⊦ ordered([], x)) Csimpnamedseq(pushx-base,,, ⊦ pushx(n, [], x) = (n + []) + []) Csimpnamedseq(ordered-single,,, ⊦ ordered(ix ', x) ↔ ordered(ix, x)) Csimpnamedseq(ordered-rec1,,, ⊦ ordered(n ', x)) Csimpgen(illist freely generated by [], +;) Csimpnamedseq(list,indexseq,c, ⊦ n + ix = n ' + ix) Csimpnamedseq(list,indexseq,app, ⊦ ix + [] = ix) Csimpnamedseq(list,indexseq,a, ⊦ (ix + ix0) + ix1 = ix + ix0 + ix1) Csimpnamedseq(list,indexseq-stackseq,c, ⊦ ix + ixs = ix ' + ixs) Csimpnamedseq(list,indexseq-stackseq,app, ⊦ ixs + [] = ixs) Csimpnamedseq(list,indexseq-stackseq,a, ⊦ (ixs + ixs0) + ixs1 = ixs + ixs0 + ixs1)))]