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Proof step 9 from proof abs-nodes-elems-adr-length-h (previous step)

abs-nodes(buf1, m + # buf0, [], lax, weak), abs-padding(buf0), ∀ buf2, m, x, lax, weak. # buf2 < #(buf0 + buf1) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ lax = []

User applied rule apply induction

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: ⊦ # [] = 0; (s)% ⊦ # lax = 0 ↔ lax = []; (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: buf1, m + # buf0, [], lax, weak