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Proof step 8 from proof del1-01 (previous step)

∀ a, b. #oc(a, x0 -1l b) = #oc(a, x0) ∨ a = b ⊦ a0 = a, a = b, #oc(a, (a0 ' + x0) -1l b) = #oc(a, x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a)
Proof step 10:
b ≠ a0, ∀ a, b. #oc(a, x0 -1l b) = #oc(a, x0) ∨ a = b ⊦ a0 = a, a = b, #oc(a, a0 ' + (x0 -1l b)) = #oc(a, x0)

Rule argument

spec: list-del inst: name: del1-n seq: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a) subst: [a, x, b] → [b, x0, a0] flags: rewrite

Additional information

Used simplifier rules: del1-y: ⊦ (a ' + x) -1l a = x; from specification list-del%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: b, x0, a0