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Proof step 4 from proof precedesex-rev-rec1 (previous step)

precedes∃([] + ix0 ' + ix1 ', x) ⊦ precedes∃([] + ix0 ', x) ∧ (n ∈ ix1 → (∃ n0. n0 ∈ ix0 ∧ precedes(n0, n, x)))

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(precedesex-2,,, ⊦ precedes∃(ix0 ' + ix1 ', x) ↔ (∀ n1. n1 ∈ ix1 → (∃ n0. n0 ∈ ix0 ∧ precedes(n0, n1, x)))) Csimpnamedseq(precedesex-rec0,,, ⊦ precedes∃(ix ', x)) Csimpnamedseq(list,indexseq-stackseq,a, ⊦ (ixs + ixs0) + ixs1 = ixs + ixs0 + ixs1) Csimpnamedseq(list,indexseq-stackseq,Nil, ⊦ [] + ixs = ixs)))]