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Proof step 9 from proof ex (previous step)

ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ ins≤(a, (a0 ' + y) + z) = (a0 ' + y) + a ' + z, ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x
Proof step 11:
ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ a + a0 + (y + z) = (a0 ' + y) + a ' + z, a0 < a, ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

Rule argument

spec: <toplevel> name: ins-y seq: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x subst: [x, b, a] → [y + z, a0, a] flags: <all paths>

Additional information

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