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Proof step 8 from proof mod-other-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
rs1 ∩ rs0 = ∅, a[P][rs1, f, g], R[rs0, f, g] ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f(r; o), g(r; a0)], a[P][rs1, f(r; o), g(r; a0)]

Rule argument

Additional information

Used simplifier rules: ⊦ (true)[rs, f, g]; (s)