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Proof step 10 from proof at-rw (previous step)

r ∈ rs ↔ r ∈ H0 ∨ r ∈ H, ((r => o))(H0), P(H), H0 ⊥ H, r ∈ rs → f(r) = (H0 ∪ H)[r] ⊦ f(r) = o

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
f(r) = H0[r], ((r => o))(H0), P(H), r ∈ rs, H0 ⊥ H ⊦ # rs = 0, H0[r] = o, rs = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]%r ∈ H0 ∪ H1 ⊦ H0 ⊥ H1 → (H0 ∪ H1)[r] = (r ∈ H0 ⊃ H0[r];H1[r]); (s)% ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)% ⊦ ((r => o))(H) → r ∈ H; (s)