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Proof step 30 from proof mod-other-both (previous step)

rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], r0 ∈ rs1, ∀ r. r ∈ rs1 → g(r) = a ⊦ # rs1 = 0, a0 = a, g(r0) = a, rs1 = ∅, (a[P] * true)[rs1 ∪ rs0, f(r0; o), g(r0; a0)]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r0

Additional information

Used terms: r0