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Proof step 7 from proof abs-nodes-prefix-elim (previous step)

abs-nodes(buf0, 0, x0, lax0, false), abs-nodes(buf0 + buf1, 0, x1, lax1, boolvar1), x0 ⊑ x1, lax0 ⊑ lax1 ⊦ ∃ x2, lax2. x1 = x0 + x2 ∧ lax1 = lax0 + lax2 ∧ abs-nodes(buf1, # buf0, x2, lax2, boolvar1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 9:
abs-nodes(buf0 + buf1, 0, x1, lax1, boolvar1), abs-nodes(buf0, 0, x0, lax0, false), lax0 ⊑ lax1, ∃ z. x0 + z = x1 ⊦ ∃ x2, lax2. x1 = x0 + x2 ∧ lax1 = lax0 + lax2 ∧ abs-nodes(buf1, # buf0, x2, lax2, boolvar1)

Rule argument

spec: list-last name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x1, x0] flags: <all paths>

Additional information

Used terms: x1, x0