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Proof step 12 from proof del1-01-02 (previous step)

ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = a ' + ((a0 ' + x0) -1l b), a0 < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a)
Proof step 14:
b = a0, ordered≤(a0 ' + x0), ∀ a1, b0. a1 = b0 ∨ ins≤(a1, x0 -1l b0) = ins≤(a1, x0) -1l b0 ⊦ a = b, ins≤(a, (a0 ' + x0) -1l b) = a ' + ((a0 ' + x0) -1l b), a0 < a

Rule argument

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

Additional information

Used simplifier rules: ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x;% ⊦ a + x = a ' + x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s) Used terms: b, x0, a0