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Proof step 18 from proof abs-nodes-unique-strong-h (previous step)

abs-nodes(buf0, m + # buf1, nd ' + x, @(m0, m1) ' + lax, false), abs-padding(buf1), ∀ buf2, m, x0, lax0, x1, lax1, boolvar1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, false) → (abs-nodes(buf2, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf0, m + # buf1, x1, lax1, boolvar1) ↔ nd ' + x = x1 ∧ @(m0, m1) ' + lax = lax1

User applied rule apply induction

And closed the goal!

Rule argument

subst: [buf2, m, x0, lax0, x1, lax1, boolvar1] → [buf0, m + # buf1, nd ' + x, @(m0, m1) ' + lax, x1, lax1, boolvar1]

Additional information

Used simplifier rules: ⊦ #(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, nd ' + x, @(m0, m1) ' + lax, x1, lax1, boolvar1