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Proof step 10 from proof abs-nodes-prefix-elim (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ lax ⊑ lax0 ↔ (∃ lax1. lax + lax1 = lax0)
Proof step 12:
abs-nodes(buf0 + buf1, 0, x0 + z, lax1, boolvar1), abs-nodes(buf0, 0, x0, lax0, false), ∃ lax2. lax0 + lax2 = lax1 ⊦ ∃ lax. lax1 = lax0 + lax ∧ abs-nodes(buf1, # buf0, z, lax, boolvar1)

Rule argument

spec: list-last, inst: laddress-list name: prefix seq: ⊦ lax ⊑ lax0 ↔ (∃ lax1. lax + lax1 = lax0) subst: [lax0, lax] → [lax1, lax0] flags: <all paths>

Additional information

Used terms: lax1, lax0