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Proof step 4 from proof from-to-exchange (previous step)

∀ 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⌝

User applied rule all left

And obtained the following new goals

Proof step 5:
(tl(σ) to m0 + n) from m0 = (tl(σ) from m0) to n ∨ # tl(σ) < ⌜m0⌝ + ⌜n⌝ , ∀ 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

pos: left 1 terms: tl(σ), n abort: F discard: F computed: F

Additional information

Used terms: tl(σ), n