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Proof step 97 from proof partitions-mod-count (previous step)

P[n] = pe(m, m0, m1), n < # P, partitions(P, n0) ⊦ P = [], partitions(P[n, pe(m, m0, m2)], n0)

User applied rule structural induction

And obtained the following new goals

Proof step 96:
∀ n, m, m0, m1, n0, m2. P0[n] = pe(m, m0, m1) ∧ n < # P0 ∧ partitions(P0, n0) → P0 = [] ∨ partitions(P0[n, pe(m, m0, m2)], n0), (p + P0)[n] = pe(m, m0, m1), n < #(p + P0), partitions(p + P0, n0) ⊦ p + P0 = [], partitions((p + P0)[n, pe(m, m0, m2)], n0)

Rule argument

indvar:P

Additional information

[Concretesimprules((Csimpgen(partition freely generated by [], +;)))]