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Proof step 36 from proof cf-map-unknown (previous step)

cf(r) = a × r0, rs1 ∩ rs0 = ∅, R[rs0, cf, af], ∀ r. r ∈ rs1 → af(r) = o, ∀ r0. (r0 ∈ mk(rs1, cf) ↔ r0 ∈ (∅ ++ r)[r, ce]) ∧ (r0 ∈ mk(rs1, cf) → mk(rs1, cf)[r0] = (∅ ++ r)[r, ce][r0]) ⊦ r = null, (o[r => a × r0] * R)[rs1 ∪ rs0, cf, af], (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → cf(r1) = (r1 = r ⊃ a × r0;(∅ ++ r)[r1]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
cf(r) = a × r0, rs1 ∩ rs0 = ∅, R[rs0, cf, af], ∀ r. r ∈ rs1 → af(r) = o, ∀ r2. (r2 ∈ rs1 ↔ r2 = r) ∧ (r2 ∈ rs1 → cf(r2) = (r2 = r ⊃ ce;(∅ ++ r)[r2])) ⊦ r = null, (o[r => a × r0] * R)[rs1 ∪ rs0, cf, af], (r1 ∈ rs1 ↔ r1 = r) ∧ (r1 ∈ rs1 → cf(r1) = (r1 = r ⊃ a × r0;(∅ ++ r)[r1]))

Rule argument

Additional information

Used simplifier rules: ⊦ H[r0, ce][r] = (r = r0 ∧ r0 ∈ H ⊃ ce;H[r]); (s)% ⊦ r ∈ rs → mk(rs, cf)[r] = cf(r); (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)% ⊦ ¬ r ∈ ∅; (s)% ⊦ r ∈ H[r0, ce] ↔ r ∈ H; (s)% ⊦ r ∈ mk(rs, cf) ↔ r ∈ rs; (s)