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Proof step 10 from proof abs-nodes-first-adr-explicit (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [buf2, m, a, x, lax, weak] → [buf0, m + # buf1, a, x, lax, weak]

Additional information

Used simplifier rules: in-14: ⊦ ladr ' + lax = lax0 → ladr ∈ lax0; (forward) from specification list[laddress-list]%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ (ladr ' + lax).head = ladr; (s)% ⊦ @(m, m0) = @(m1, m2) ↔ m = m1 ∧ m0 = m2; (ws)% ⊦ ladr ∈ ladr0 ' + lax ↔ ladr = ladr0 ∨ ladr ∈ lax; (s)% ⊦ ladr ' + lax = ladr ' + lax0 ↔ lax = lax0; (s)% ⊦ #(buf + buf0) = # buf + # buf0; (s)%append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ # buf = 0 ↔ buf = mkbuf(0); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ # mkbuf(n) = n; (s)% ⊦ n + 0 = n; (s)% ⊦ ¬ abs-padding(mkbuf(0)); (s) Used terms: buf0, m + # buf1, a, x, lax, weak