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Proof step 3 from proof countrefs-lastn-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m + 1 ≤ n ↔ m < n], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n])]]