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Proof step 5 from proof extensionality-to (previous step)

# σ0 = # σ1, ⌜n⌝ < # σ1, ∀ n. ¬ # σ1 < ⌜n⌝ → σ0 to n = σ1 to n ⊦ σ0[n] = σ1[n]

User applied rule all left

And obtained the following new goals

Proof step 6:
# σ0 = # σ1, ⌜n⌝ < # σ1, ¬ # σ1 < ⌜n + 1⌝ → σ0 to n + 1 = σ1 to n + 1, ∀ n. ¬ # σ1 < ⌜n⌝ → σ0 to n = σ1 to n ⊦ σ0[n] = σ1[n]

Rule argument

pos: left 3 terms: n + 1 abort: F discard: F computed: F

Additional information

Used terms: n + 1