Theorem example-E7.1a 1074

Description: '𝑥' is conditionally not free in the statement '(𝑥 ⋅ 𝑦) = 0' when '𝑦 = 0'. †

Since no arithmetic axioms have been introduced, we state '(𝑥 ⋅ 0) = 0' as a hypothesis.

Hypothesis

Ref	Expression
example-E7.1a.1	⊢ (𝑥 ⋅ 0) = 0

Assertion

Ref	Expression
example-E7.1a	⊢ (𝑦 = 0 → Ⅎ𝑥 (𝑥 ⋅ 𝑦) = 0)

Distinct variable group:   𝑥,𝑦

Proof of Theorem example-E7.1a

Step	Hyp	Ref	Expression
1		ndnfrv-P7.1 826	. . 3 ⊢ Ⅎ𝑥 0 = 0
2	1	rcp-NDIMP0addall 207	. 2 ⊢ (𝑦 = 0 → Ⅎ𝑥 0 = 0)
3		ndsubmultr-P7.24e.CL 922	. . . . 5 ⊢ (𝑦 = 0 → (𝑥 ⋅ 𝑦) = (𝑥 ⋅ 0))
4		example-E7.1a.1	. . . . . 6 ⊢ (𝑥 ⋅ 0) = 0
5	4	rcp-NDIMP0addall 207	. . . . 5 ⊢ (𝑦 = 0 → (𝑥 ⋅ 0) = 0)
6	3, 5	eqtrns-P7 987	. . . 4 ⊢ (𝑦 = 0 → (𝑥 ⋅ 𝑦) = 0)
7	6	ndsubeql-P7.22a 847	. . 3 ⊢ (𝑦 = 0 → ((𝑥 ⋅ 𝑦) = 0 ↔ 0 = 0))
8	7	ndnfrleq-P7.11.VR 862	. 2 ⊢ (𝑦 = 0 → (Ⅎ𝑥 (𝑥 ⋅ 𝑦) = 0 ↔ Ⅎ𝑥 0 = 0))
9	2, 8	bimpr-P4 533	1 ⊢ (𝑦 = 0 → Ⅎ𝑥 (𝑥 ⋅ 𝑦) = 0)

Syntax hints:  term-obj 1  0term_zero 2   ⋅ term-mult 5   = wff-equals 6   → wff-imp 10  Ⅎwff-nfree 681

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14  ax-GEN 15  ax-L4 16  ax-L5 17  ax-L6 18  ax-L7 19  ax-L9-multr 26  ax-L12 29

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161  df-exists-D5.1 596  df-nfree-D6.1 682

This theorem is referenced by: (None)