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Proof step 5 from proof mod-other-both (previous step)

∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ a[P][rs0, f, g] ∧ R[rs1, f, g] ⊦ g(r) = a, (a[P] * true)[rs, f(r; o), g(r; a0)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information