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Proof step 5 from proof write-other-true (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
rs1 ∩ rs0 = ∅, a[P][rs1, f(r; o), g] ⊦ g(r) = a, (a[P] * true)[rs1 ∪ rs0, f, g]

Rule argument

Additional information

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