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Proof step 87 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, P(mk(rs2, f0)), r ∈ rs2, ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)), ∀ r. r ∈ rs2 → g0(r) = a ⊦ # rs2 = 0, g(r) = a, rs2 = ∅

User applied rule all left

And obtained the following new goals

Proof step 88:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1, rs2 ∩ rs1 = ∅, P(mk(rs2, f0)), r ∈ rs2, g(r) = a ∧ r ∈ rs → f(r) = f0(r), ∀ r. r ∈ rs2 → g0(r) = a, ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ # rs2 = 0, g(r) = a, rs2 = ∅

Rule argument

pos: left 4 terms: r abort: F discard: F computed: T

Additional information

Used terms: r