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Proof step 9 from proof abs-nodes-first-adr-explicit (previous step)

lax.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, a ' + x, lax, weak) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, a) ∧ abs-nodes(buf0, m + # buf1, x, lax.tail, weak), ∀ buf2, m, a, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, a ' + x, lax, weak) → (∃ lax0. lax = lax.head ' + lax0) ⊦ ∃ lax0. lax = @(m0, m1) ' + lax0

User applied rule case distinction

And obtained the following new goals

Proof step 10:
lax.head = @(m0, m1), abs-nodes(buf0, m + # buf1, a ' + x, lax, weak), abs-padding(buf1), ∀ buf2, m, a, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, a ' + x, lax, weak) → (∃ lax0. lax = lax.head ' + lax0) ⊦ ∃ lax0. lax = @(m0, m1) ' + lax0
Proof step 11:
lax.head = @(m0, m1), m0 = m, m1 = # buf1, abs-node(buf1, a), abs-nodes(buf0, m + # buf1, x, lax.tail, weak), ∀ buf2, m, a, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, a ' + x, lax, weak) → (∃ lax0. lax = lax.head ' + lax0) ⊦ lax = [], ∃ lax0. lax = @(m0, m1) ' + lax0

Rule argument

pos: left 2

Proof step 9 from proof abs-nodes-first-adr-explicit (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information