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Proof step 4 from proof abs-nodes-prefix-elim (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
⊦ abs-nodes(buf0, m, x0, lax0, boolvar0) ∧ abs-nodes(buf0 + buf1, m, x1, lax1, boolvar1) → x0 ⊑ x1 ∧ lax0 ⊑ lax1
Proof step 6:
abs-nodes(buf0 + buf1, 0, x1, lax1, boolvar1), abs-nodes(buf0, 0, x0, lax0, false), 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)

Rule argument

spec: <toplevel> name: abs-nodes-prefix-h seq: ⊦ abs-nodes(buf0, m, x0, lax0, boolvar0) ∧ abs-nodes(buf0 + buf1, m, x1, lax1, boolvar1) → x0 ⊑ x1 ∧ lax0 ⊑ lax1 subst: [buf0, m, x0, lax0, boolvar0, buf1, x1, lax1, boolvar1] → [buf0, 0, x0, lax0, false, buf1, x1, lax1, boolvar1]

Additional information

Used terms: buf0, 0, x0, lax0, false, buf1, x1, lax1, boolvar1