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Proof step 17 from proof resetcnt-mod-cnt (previous step)

∀ n, m, m0, m1, m2. P0[n] = pe(m, m0, m1) ∧ n < # P0 → resetcnt(P0[n, pe(m, m0, m2)]) = resetcnt(P0) ∨ P0 = [], (p + P0)[n] = pe(m, m0, m1), n < #(p + P0) ⊦ resetcnt((p + P0)[n, pe(m, m0, m2)]) = resetcnt(p + P0), p + P0 = []

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

[Concretesimprules((Csimpnamedseq(resetcnt-rec,,, ⊦ resetcnt(p ' + P) = p.count:= 0 + resetcnt(P)) Csimpnamedseq(nat,,lels, ⊦ ¬ m < n ↔ n ≤ m) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic2,,s, ⊦ m < n + 1 ↔ ¬ n < m) Csimpnamedseq(list,partition,len-02, ⊦ # p ' = 0 +1) Csimpnamedseq(list,partition,len, ⊦ #(P + P0) = # P + # P0) Csimpnamedseq(list,partition,e-01, ⊦ p ' + P ≠ []) Csimpnamedseq(list,partition,a, ⊦ (P + P0) + P1 = P + P0 + P1) Csimpnamedseq(list,partition,c, ⊦ p + P = p ' + P)))]