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Proof step 1 from proof exencode (conclusion) (back to tree)

⊦ ∃ i, j. i × j = i0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i0.1, i0.2 abort: F discard: T computed: F

Additional information

Used simplifier rules: dec-fst: ⊦ i.1 = decode(decode(i).1); dec-snd: ⊦ i.2 = decode(decode(i).2); encode-pair: ⊦ i × j = encode(encode(encode(i), encode(j))); dec-nat-01: ⊦ encode(decode(n)) = n; dec-int-01: ⊦ encode(decode(i)) = i; ⊦ encode(m.1, m.2) = m; (s) Used terms: i0.1, i0.2

Proof Tree for exencode (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree