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Proof step 21 from proof sep (previous step)

rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦ r ∈ rs0 ∪ rs1 ↔ r ∈ rs

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ r ∈ rs ∪ rs0 ↔ ¬(¬ r ∈ rs ∧ ¬ r ∈ rs0); (ss)%una: ⊦ (rs ∪ rs0) ∪ rs1 = rs ∪ rs0 ∪ rs1; from specification set-union[refset]%unc: ⊦ rs ∪ rs0 = rs0 ∪ rs; from specification set-union[refset]% ⊦ ¬ r ∈ rs → (r ∈ rs ∪ rs0 ↔ r ∈ rs0); (s) Used terms: r