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Proof step 55 from proof rely-owned-unchgd (previous step)

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)) ⊦ rs2 ∪ (rs \ rs2) = rs ∧ a[P][rs2, f, g]

User applied rule case distinction

And obtained the following new goals

Proof step 56:
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)) ⊦ rs2 ∪ (rs \ rs2) = rs
Proof step 68:
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][rs2, f, g]

Rule argument

pos: right 1

Proof step 55 from proof rely-owned-unchgd (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information