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Proof step 95 from proof partitions-mod-count (previous step)

(p ' + P0)[n] = pe(m, m0, m1), partitions(p ' + P0, n0), n ≤ # P0, ∀ 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) ⊦ partitions((p ' + P0)[n, pe(m, m0, m2)], n0)

User applied rule cut formula

And obtained the following new goals

Proof step 37:
n = 0, (p ' + P0)[n] = pe(m, m0, m1), partitions(p ' + P0, n0), n ≤ # P0, ∀ 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) ⊦ partitions((p ' + P0)[n, pe(m, m0, m2)], n0)

Rule argument

fma:n = 0

Additional information

[Concretesimprules(()), Extrafmas((n = 0))]