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Proof step 1 from proof enc-zero (conclusion)

⊦ i × j = 0 ↔ ¬ ¬(i = 0 ∧ j = 0)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: encode-pair: ⊦ i × j = encode(encode(encode(i), encode(j)));%enc-zero-02: ⊦ encode(n) = 0 ↔ n = 0;%enc-int-pos: ⊦ 0 ≤ i → encode(i) = 2 * i→n(i);%enc-zero-01: ⊦ encode(i) = 0 ↔ i = 0;%encode-zero: ⊦ encode(m, n) = 0 ↔ m = 0 ∧ n = 0; from specification nat-pair%encode-zero-01: ⊦ encode(0, 0) = 0; from specification nat-pair%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult