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Proof step 49 from proof rely-owned-unchgd (previous step)

∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)), ∃ rs2, rs1. rs2 ∪ rs1 = rs0 ∧ rs2 ∩ rs1 = ∅ ∧ a[P][rs2, f0, g0] ∧ (true)[rs1, f0, g0] ⊦ (a[P] * true)[rs, f, g]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
rs2 ∩ rs1 = ∅, a[P][rs2, f0, g0], ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (a[P] * true)[rs, f, g]

Rule argument

Additional information

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