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Proof step 3 from proof member-reset (previous step)

∀ m, m0, m1. pe(m, m0, m1) ∈ P0 → pe(m, m0, 0) ∈ resetcnt(P0), pe(m, m0, m1) ∈ p + P0 ⊦ pe(m, m0, 0) ∈ resetcnt(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(partelem,,upd-03, ⊦ pe(m, m0, m1) .count:= m4 = pe(m, m0, m4)) Csimpnamedseq(list,partition,in-06, ⊦ ¬ p ∈ p0 ' + P ↔ p ≠ p0 ∧ ¬ p ∈ P) Csimpnamedseq(list,partition,a, ⊦ (P + P0) + P1 = P + P0 + P1) Csimpnamedseq(list,partition,in-17, ⊦ p ∈ p0 ' ↔ p = p0) Csimpnamedseq(list,partition,in-append, ⊦ p ∈ P + P0 ↔ ¬ ¬(p ∈ P ∨ p ∈ P0)) Csimpnamedseq(list,partition,c, ⊦ p + P = p ' + P)))]