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Proof step 11 from proof subsize (previous step)

a ∈ y, #oc(a, x0) < #oc(a, y), # y ≤ # x0, ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 13:
∃ y0, z. y = y0 + (a + z), #oc(a, x0) < #oc(a, y), # y ≤ # x0, ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦ x0 = [], ∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))

Rule argument

spec: list inst: name: In seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)) subst: [x, a] → [y, a] flags: rewrite

Additional information

Used terms: y, a