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Proof step 2 from proof subrec-01 (previous step)

∀ a, y. x0 ⊆m a ' + y ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y ⊦ a0 + x0 ⊆m a ' + y ↔ #oc(a, a0 + x0) ≤ #oc(a, y) + 1 ∧ (a0 + x0) -1l a ⊆m y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a, y. x0 ⊆m a ' + y ↔ #oc(a, x0) ≤ #oc(a, y) + 1 ∧ x0 -1l a ⊆m y ⊦ (a0 = a ∨ a0 ∈ y) ∧ x0 ⊆m (a ' + y) -1l a0 ↔ #oc(a, a0 ') + #oc(a, x0) ≤ #oc(a, y) + 1 ∧ (a0 ' + x0) -1l a ⊆m y

Rule argument

Additional information

Used simplifier rules: subrec: ⊦ a ' + x ⊆m y ↔ a ∈ y ∧ x ⊆m y -1l a;%ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)