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Proof step 8 from proof ocs-del-length (previous step)

∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ a = a0, a = b, #oc(a, (a0 ' + x0) -l b) = #oc(a, x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a)
Proof step 10:
¬ b ≠ a0, ∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ a = a0, a = b, #oc(a, (a0 ' + x0) -l b) = #oc(a, x0)
Proof step 12:
b ≠ a0, ∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ a = a0, a = b, #oc(a, a0 ' + (x0 -l b)) = #oc(a, x0)

Rule argument

spec: list-del inst: name: del-n seq: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a) subst: [a, x, b] → [b, x0, a0] flags: rewrite

Additional information

Used terms: b, x0, a0