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Proof step 5 from proof rely-trans (previous step)

g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, g0(r) = a ∧ r ∈ rs0 → f(r) = f0(r), ∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)), ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f(r) = f1(r))

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: r

Additional information

Used simplifier rules: ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset] Used terms: r