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Proof step 5 from proof countrefs-lastn-02 (conclusion)

⊦ r ∈ refs1 ∧ ¬ r ∈ refs ∧ # refs = 2 + n * m + m0 → lastn(1 + n * m, refs) ≠ r0 ' + refs1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
lastn(1 + n * m, refs) = r0 ' + refs1, # refs = 2 + n * m + m0, r ∈ refs1 ⊦ r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n])]]