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Proof step 3 from proof from-to-exchange (previous step)

⊦ (σ to m + n) from m = (σ from m) to n, # σ < ⌜m⌝ + ⌜n⌝

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ σ, n. (σ to m0 + n) from m0 = (σ from m0) to n ∨ # σ < ⌜m0⌝ + ⌜n⌝ ⊦ (σ to m0 +1 + n) from m0 +1 = (σ from m0 +1) to n, # σ < ⌜m0 +1⌝ + ⌜n⌝

Rule argument

indvar: m

Additional information

Used simplifier rules: from-zero: ⊦ σ from 0 = σ;%nat freely generated by 0, +1;;% ⊦ 0 + m = m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic