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Proof step 43 from proof swap-to-from-app (previous step)

#((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, #((σ to m + n0) from m) = ⌜n1⌝ , ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, n0 ≤ n1 + n2 ⊦ σ0 = [], σ0[n2 + n1 - n0] = σ0[n2], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 44:
n2 + n1 = n3 + n0 ∧ n0 ≤ n2 + n1, #((σ to n0 + m) + σ0 from m) = ⌜n1⌝ + # σ0, #((σ to m + n0) from m) = ⌜n1⌝ , ⌜n1⌝ < ⌜n0⌝ + # σ0, ⌜n2⌝ < # σ0, ⌜n2⌝ + ⌜n1⌝ < ⌜n0⌝ + # σ0, n0 ≤ n1 + n2 ⊦ σ0 = [], σ0[n3] = σ0[n2], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n2 + n1, n0, n] flags:

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2 + n1, n0, n