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Proof step 15 from proof mod-other-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
rs1 ∩ rs0 = ∅, P(mk(rs1, f)), R[rs0, f, g], ∀ r. r ∈ rs1 → g(r) = a ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)], a[P](mk(rs1, f(r; o), g(r; a0)))

Proof step 15 from proof mod-other-both (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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