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Proof step 5 from proof del1-02 (previous step)

a0 < a, ∀ a. ins≤(a, x0) -1l a = x0 ⊦ (a0 ' + ins≤(a, x0)) -1l a = a0 ' + x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a)
Proof step 7:
a ≠ a0, a0 < a, ∀ a. ins≤(a, x0) -1l a = x0 ⊦ a0 ' + (ins≤(a, x0) -1l a) = a0 ' + 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] → [a, ins≤(a, 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%irreflexivity: ⊦ ¬ a < a; from specification oelem Used terms: a, ins≤(a, x0), a0