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Proof step 7 from proof range-eq-append (previous step)

∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ range=(ar0, n0, ar1, n1, (m + 1) + m1) ↔ range=(ar0, n0, ar1, n1, m + 1) ∧ range=(ar0, n0 + m + 1, ar1, n1 + m + 1, m1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m)
Proof step 9:
∀ ar0, n0, ar1, n1, m1. range=(ar0, n0, ar1, n1, m + m1) ↔ range=(ar0, n0, ar1, n1, m) ∧ range=(ar0, n0 + m, ar1, n1 + m, m1) ⊦ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m1 + m) ↔ range=(ar0, n0, ar1, n1, m + 1) ∧ range=(ar0, n0 + m + 1, ar1, n1 + m + 1, m1)

Rule argument

spec: <toplevel> name: range-eq-rec seq: ⊦ range=(ar0, n0, ar1, n1, m + 1) ↔ ar0[n0] = ar1[n1] ∧ range=(ar0, n0 + 1, ar1, n1 + 1, m) subst: [m, n1, ar1, n0, ar0] → [m1 + m, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m1 + m, n1, ar1, n0, ar0