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Proof step 20 from proof abs-nodes-unique-weak-h (previous step)

abs-nodes(buf, (m + # buf0) + # buf2, [], [], true), abs-nodes(buf, (m + # buf0) + # buf2, nd ' + x2, @(m0, m1) ' + lax, true), abs-padding(buf0), abs-padding(buf2), ∀ buf3, m, x0, lax0, x1, lax1. # buf3 < #(buf0 + buf2 + buf) → abs-nodes(buf3, m, x0, lax0, true) ∧ abs-nodes(buf3, m, x1, lax1, true) → x0 = x1 ∧ lax0 = lax1 ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [buf3, m, x0, lax0, x1, lax1] → [buf, (m + # buf0) + # buf2, nd ' + x2, @(m0, m1) ' + lax, [], []]

Additional information

Used simplifier rules: ⊦ lax + lax0 = [] ↔ lax = [] ∧ lax0 = []; (as)% ⊦ ladr ' ≠ []; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ #(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)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ # buf = 0 ↔ buf = mkbuf(0); (s)% ⊦ # mkbuf(n) = n; (s)% ⊦ n + 0 = n; (s)% ⊦ ¬ abs-padding(mkbuf(0)); (s) Used terms: buf, (m + # buf0) + # buf2, nd ' + x2, @(m0, m1) ' + lax, [], []