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Proof step 61 from proof splice-resize-less (previous step)

# ar1 < n1 + n2, max(# ar1, n1 + m0) < n2 + n1, n3 < n2, n3 < m0, max(# ar1, n1 + m0) ≤ n3 + n1, n2 ≤ m0 ⊦ zero = ar0[n3 + n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: min-max name: max seq: ⊦ max(n, m) = (n < m ⊃ m;n) subst: [m, n] → [n1 + m0, # ar1] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: n1 + m0, # ar1