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Proof step 3 from proof inv-last-nonempty (previous step)

∀ n. ixs0.last = [] ∧ inv(ixs0, n) → ixs0 = [], (ix + ixs0).last = [], inv(ix + ixs0, n) ⊦ ix + ixs0 = []

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(inv-rec,,, ⊦ inv(ix ' + ixs, n) ↔ ix ≠ [] ∧ all<(ix, n) ∧ inv(ixs, n)) Csimpnamedseq(list-last,indexseq-stackseq,l-02, ⊦ ixs ≠ [] → (ix ' + ixs).last = ixs.last) Csimpnamedseq(list-last,indexseq-stackseq,l, ⊦ ix '.last = ix) Csimpnamedseq(list,indexseq-stackseq,app, ⊦ ixs + [] = ixs) Csimpnamedseq(list,indexseq-stackseq,a, ⊦ (ixs + ixs0) + ixs1 = ixs + ixs0 + ixs1) Csimpnamedseq(list,indexseq-stackseq,e-01, ⊦ ix ' + ixs ≠ []) Csimpnamedseq(list,indexseq-stackseq,c, ⊦ ix + ixs = ix ' + ixs)))]